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Summer 2019 INTRODUCTION OF INFINITE SERIES IN HIGH SCHOOL LEVEL CALCULUS Ericka Bella John Carroll University, [email protected]
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Recommended Citation Bella, Ericka, "INTRODUCTION OF INFINITE SERIES IN HIGH SCHOOL LEVEL CALCULUS" (2019). Masters Essays. 122. https://collected.jcu.edu/mastersessays/122
This Essay is brought to you for free and open access by the Master's Theses and Essays at Carroll Collected. It has been accepted for inclusion in Masters Essays by an authorized administrator of Carroll Collected. For more information, please contact [email protected]. INTRODUCTION OF INFINITE SERIES IN HIGH SCHOOL LEVEL CALCULUS
An Essay Submitted to the Office of Graduate Studies College of Arts & Sciences of John Carroll University In Partial Fulfillment of the Requirements For the Degree of Master of Arts
By Ericka L. Bella 2019
The essay of Ericka L. Bella is hereby accepted:
______
Advisor – Barbara K. D’Ambrosia Date
I certify that this is the original document
______
Author – Ericka L. Bella Date Table of Contents
Introduction ...... 1
Lesson 1: Introduction to Sequences and Series Teacher Notes ...... 3 Guided Notes with Answers ...... 4 Activity Sheet with Answers ...... 7
Lesson 2: The nth-Term Test for Divergence Teacher Notes ...... 10 Guided Notes with Answers ...... 11 Activity Sheet with Answers ...... 17
Lesson 3: Geometric Series Teacher Notes ...... 21 Guided Notes with Answers ...... 22 Activity Sheet with Answers ...... 28
Lesson 4: p-Series Teacher Notes ...... 34 Guided Notes with Answers ...... 35 Activity Sheet with Answers ...... 45
Lesson 5: Direct Comparison Test Teacher Notes ...... 49 Guided Notes with Answers ...... 50 Activity Sheet with Answers ...... 59
Lesson 6: Limit Comparison Test Teacher Notes ...... 63 Guided Notes with Answers ...... 64 Activity Sheet with Answers ...... 74
Lesson 7: Ratio Test Teacher Notes ...... 78 Guided Notes with Answers ...... 79 Activity Sheet with Answers ...... 89
Appendix: Guided Notes and Activity Sheets for Students ...... 95
References ...... 181
Introduction
For the past six years, I have taught a variety of calculus courses, ranging from AP Calculus BC, College Credit Plus Calculus 1 and 2, and Calculus, which is an introductory course that does not lead to college credit. The curriculum in each of these courses differs significantly. While one course starts the year with limits and differentiation, another slowly builds up to calculus topics, beginning with a hearty review of pre-calculus material. Depending on which calculus course students decide to take, they may end the year with derivatives and integration of trigonometric functions, while others are learning various techniques of integration.
Regardless of what calculus course my students decide to take, I want them to be as well prepared as possible for the mathematics classes they may take in college. For most of my students, calculus is the last mathematics class they will take in high school and I believe exposing them to new content is essential, even if we cannot cover it in depth due to time restrictions. Many of my past students have come back to visit while in college and mentioned that learning calculus topics in college has been significantly easier because they have had some exposure to these ideas in high school.
For this essay, I have designed an introduction to sequences and series for my non- college level Calculus course. This course begins with precalculus material and ends with the calculus of trigonometric functions. The students in this course do not encounter any content regarding sequences or series in the current curriculum. Based on my previous experiences as a student, entering college calculus without having learned about the convergence and divergence of series was a challenging task. There are a multitude of tests and rules to learn and knowing when and how to apply each can be confusing.
This introductory unit includes the following for each lesson: teacher notes, guided notes sheets, and an activity sheet, with solutions printed in red. Blank student documents are available in the appendix. I used language, notation, and organization of the sequences and series material as in the textbook Calculus of a Single Variable by Larson, Hostetler, and Edwards [2].
Due to the daily schedule at my school, I have planned each lesson for an 80-minute class period. I will lead students through the guided notes, with students completing the notes as reference for future use. Students will work in groups of two or three to complete the activity for the lesson. Students will provide their own technology for each lesson, as all students in my courses are required to own a TI-Nspire CX calculator.
Since there are no Ohio Mathematics Learning Standards for Calculus, I have linked standards for each activity to the AP Calculus BC College Board course description.
1 Additionally, all activities reflect the following four mathematical practices as defined by College Board [1].
Mathematical Practice 1: Determine expressions and values using mathematical procedures and rules.
Mathematical Practice 2: Translate mathematical information from a single representation or across multiple representations.
Mathematical Practice 3: Justify reasoning and solutions.
Mathematical Practice 4: Use correct notation, language, and mathematical conventions to communicate results or solutions.
Overall, I hope to expose my high school calculus students to the foundation of sequences and series. This is not intended to be an all-encompassing study, but instead an introduction on which they can later build. Although there are many tests for determining the convergence of series, I have purposefully selected six in particular to focus on: The nth-Term Test, Geometric Series, p-Series, Direct Comparison Test, Limit Comparison Test, and Ratio Test. Students do not learn partial fraction decomposition or infinite integrals in this course. Therefore, I have decided to leave Telescoping Series and the Integral Test out of this unit.
2 Lesson 1: Introduction to Sequences and Series
Teacher Notes
Overview:
In this lesson, students will be introduced to sequences and series. Although some students enter calculus with prior knowledge of these terms, most students need a refresher to jog their memories. Although sequences and series are discussed in Algebra 2, the focus is often only arithmetic and geometric sequences and series. While these types of sequences and series are revisited in calculus, arithmetic and geometric sequences and series are very simplistic compared to what students see in a calculus course. This introduction is a great time to review factorials as well.
In this first activity, students will classify sequences and series. Students will be given sequences and asked to list terms, as well as find an expression for the nth term of a sequence. Students will predict whether a sequence will have a limit given the pattern for the nth term. Finally, students will be asked to find partial sums of a series.
When creating the examples for this activity, I wanted to focus on sequences and series that combined different types of expressions, as well as a mixture of positive and negative terms. It has been my experience that students especially struggle with finding a pattern for the nth term when the expressions for the first few terms have been simplified, so I decided to include some examples of this scenario as well.
Objectives:
• Identify sequences and series • List the terms of a sequence • Write a formula for the nth term of a sequence • Determine whether a sequence converges or diverges • Calculate partial sums of a series
Standards:
II. Limits: LIM-7.A.1: The nth partial sum is defined as the sum of the first n terms of a series.
3 Guided Notes with Answers
Introduction to Sequences and Series Name: ______
Definitions
Sequence: a function whose domain is the set of positive integers; for the purpose of this course, a sequence will refer to an infinite list of numbers that often follows a pattern
We use subscript notation to represent the terms of the sequence: aaaa123, , , ..., , ... n
th The n term is denoted by an .
The entire sequence is denoted by an .
Series: the sum of the terms of a sequence
We can calculate partial sums as well as some infinite sums.
Partial sums are denoted by Sm , where the first m terms of the sequence are added.
m Notation using sigma: Saaaaamnm==++++ 123 n=1
If the limit L of a sequence exists as n goes to infinity, then the sequence converges to L. If the limit of a sequence does not exist as n goes to infinity, then the sequence diverges.
3n Example 1: Write the first 5 terms of the sequence whose nth term is a = . n n!
331 392 32793 381274 a === 3 a == a === a === 1 1!1 2 2! 2 3 3!62 4 4!248
3243815 a === 5 5!12040
9 9 27 81 The first five terms of the sequence are 3, , , , . 2 2 8 40
4 Example 2: Find an expression for the nth term of the sequence 3 , 7 , 1 1 , 1 5 , . . . .
ann =−41
111 Example 3: Find an expression for the nth term of the sequence 1, ,−− , , ... . 2624
( 1− ) n+1 a = n n!
Example 4: Determine if the sequence converges or diverges by finding the limit, if 1 possible: a =−5. n n3
1 lim55−=3 n→ n
Since the limit exists, the sequence converges to 5.
Example 5: Determine if the sequence converges or diverges by finding the limit, if 2 possible: an = cos. n
2 lim cos1= n→ n
Since the limit exists, the sequence converges to 1.
Example 6: Determine if the sequence converges or diverges by finding the limit, if n possible: an =1 +( − 1) .
n lim 1+−( 1) does not exist. The terms of this sequence alternate between 0 and 2. n→
The sequence diverges.
5 Example 7: Determine if the sequence converges or diverges by finding the limit, if possible: ann = sin ( ) .
If we consider the function f x( x ) sin= ( ) whose domain is the set of all real numbers, then limlimsinfxx( ) = ( ) does not exist. nn→→
However, the domain of the sequence g( n )= sin ( n) is just the set of all positive integers. Therefore gn( ) 0 = for every n, and thus limg( n) == lim sin( n) 0 . nn→ →
The sequence converges.
3n Example 8: Find the partial sum S given a = . 4 n n + 2
3 63 9 12 a ==1 a == a = a ==2 1 3 2 42 3 5 4 6
3 9 63 S =12 + + + = 4 2 5 10
6 Activity Sheet with Answers
Activity: Introduction to Sequences and Series Name: ______
1. Classify each of the following as a sequence or a series.
1 1 1 1 4 8 16 32 a) , , , , ... b) + + + + 2345 5 125 625 3125
Sequence Series
12624 3333 c) ++++ d) −−, , , , ... 7777 8163264
Series Sequence
2. Create your own pattern for a sequence. Write the expression for the nth term, using correct notation. Then list the first five terms of your sequence.
n 1 Answers will vary. Sample answer: an = 4
1 1 1 1 1 a = a = a = a = a = 1 4 2 16 3 64 4 256 5 1024
3. For each of the following, write an expression for the nth term of the sequence.
5251256253125 (−5)n a) −−−, , , , , ... a = 33333 n 3
27 9 9 9 27 b) 27, , , , , ... a = 2 2 8 40 n n!
7 c) 1, 6, 11, 16, 21, ... ann =−54
2624120720 (n +1!) d) −−−−−, , , , , ... a =− 749343240116807 n 7n
Find the first five terms of each of the following sequences. Then determine if the sequence converges or diverges. If the sequence converges, find its limit.
5n 4. a = n 3+ n
5 5 20 25 a = a = 2 a = a = a = 1 4 2 3 2 4 7 5 8
5n The sequence converges: lim5. = n→ 3+ n
n 5. ann =−( 1!)
a1 =−1 a2 = 2 a3 =−6 a4 = 24 a5 =−120
Notice that the absolute value of the terms is growing without bound and the terms alternate between negative and positive. The sequence diverges and the limit does not exist.
8 5n 6. a = n n3
25 125 625 a = 5 a = a = a = a = 25 1 2 8 3 27 4 64 5
Notice that the terms are growing without bound and this can be verified with L’Hospital’s Rule. The sequence diverges and the limit does not exist.
21 7. Find the partial sum S6 of the series −+−+−+−+42( ) 36
−4 a = n n!
2 1 1 1 a =−4 a =−2 a =− a =− a =− a =− 1 2 3 3 4 6 5 30 6 180
21111237 S7 = −42 +( −+) −+ −+ −+ −= − 3630180180
9 Lesson 2: The nth-Term Test for Divergence
Teacher Notes
Overview:
Following the previous lesson, students review the topic of series at the start of this lesson. We then transition to learning about the first test we will study, the nth-Term Test for Divergence. At this point, students should be reminded that we will focus mainly on series for the remainder of this unit.
In this activity, students will find the limit of the nth term through two different methods. Students will find the limit of the nth term by algebraic means as well as complete a table of values to approximate the limit of the nth term. Students will make conclusions about whether a series diverges based on the value of this limit.
Based on my previous experience teaching this topic, I know students struggle with the conclusions that can accurately be made with the nth-Term Test. Many of my students incorrectly conclude that a series converges if lim0an = . I specifically designed the first n→ few questions of this activity to get students thinking about what this test actually means and what conditions are needed to conclude that a series diverges with this test.
Objectives:
• Find the limit of the nth term of a series • Apply the nth-Term Test for Divergence
Standards:
II. Limits: LIM-7.A.5: The nth-Term Test is a test for divergence of a series.
10 Guided Notes with Answers
The nth-Term Test for Divergence Name: ______
Recall from previous lesson
Series: the sum of the terms of a sequence
We can calculate partial sums as well as some infinite sums.
m Notation using sigma: Saaaaamnm==++++ 123 n=1
Example 1: Consider the series 0 0++++ 0 0 .
th If we let an represent the n term of this series, what is l im an ? n→
l im 0an = n→
If we added infinitely many terms of this series, what do you think the sum would be?
Zero
Based on our answers above, do you think this series converges or diverges?
Converges
111111111 Example 2: Consider the series 1 ++++++++++. 223334444
If we let represent the nth term of this series, what is ?
If we added infinitely many terms of this series, what do you think the sum would be?
The sum would grow without bound. We can group terms to add to 1 and then add infinitely many 1’s:
1 1 1 1 1 1 1 1 1 1+ + + ++ + +++ +=++++ 1 1 1 1 . 2 2 3 3 3 4 4 4 4
11 Based on our answers above, do you think this series converges or diverges?
Diverges
Example 3: Consider the series 5791113+++++ .
th If we let an represent the n term of this series, what is l im an ? n→
lim an = n→
Additionally, l im 0.an n→
If we added infinitely many terms of this series, what do you think the sum would be?
The sum would grow without bound.
Based on our answers above, do you think this series converges or diverges?
Diverges
11111 Example 4: Consider the series +++++ . 33333
If we let represent the nth term of this series, what is ?
1 lim an = n→ 3
Additionally, lim0.an n→
If we added infinitely many terms of this series, what do you think the sum would be?
The sum would grow without bound.
Based on our answers above, do you think this series converges or diverges?
Diverges
12 Based on our examples, what conclusion can we make regarding l im an and the n→ convergence or divergence of a series?
When l im 0an , the series diverges. n→
This leads us to our first test to determine whether a series diverges.
Think about the following:
If a series converges, the limit of its nth term must be 0.
That is, if a converges, then l im 0an = . n n→ n=1
It is generally easier to determine the limit of the nth term than it is to determine whether a series converges. Therefore, we will use the contrapositive of the statement above.
The nth-Term Test for Divergence
If liman 0 , then an diverges. n→
Note: If , we cannot make any conclusion about the series. Look back at Example 1 and Example 2 to verify this.
Additionally, notice that when we express a sum as an without bounds, we always mean an infinite sum, starting at a finite value of n. Since the series has infinitely many terms, the convergence or divergence of the series does not depend on the particular finite value of n we start with. However, the actual value of a convergent sum will depend on the initial value of n.
13 a a For example, let’s look at aaaann=++12 . Since 1 and 2 are finite, an nn==13 n=1 converges if and only if an converges. n=3
Example 5: Use the nth-Term Test for Divergence to determine whether the series diverges.
5 1( .2 3 )n n=0
n n an = 5 1.23( ) and limlim51.23an == ( ) nn→→
n Therefore, liman 0 and 51.23( ) diverges. n→ n=0
Example 6: Use the nth-Term Test for Divergence to determine whether the series diverges.
7 n=3 nn( + 8)
7 7 an = and limlim0an == nn( + 8) nn→→ nn( +8)
7 Since liman = 0 , we cannot make a conclusion about whether diverges. The n→ n=3 nn( + 8) nth-Term Test is inconclusive for this series.
14 Example 7: Use the nth-Term Test for Divergence to determine whether the series diverges.
n n=1 23n +
n n 1 an = and liman == lim 23n + nn→ → 2n + 3 2
n Therefore, liman 0 and diverges. n→ n=1 23n +
Example 8: Use the nth-Term Test for Divergence to determine whether the series diverges.
3n n=1 n!
3n 3n an = and limlim0an == n! nn→→ n!
3n Note: We can explain that lim0 = using the “Law of Dominance”. Although the n→ n! numerator and denominator are growing without bound, factorials grow at a faster rate than exponentials. We have previously discussed this concept in the Calculus course.
Therefore for terms far out in the sequence, the denominator is much larger than the numerator, leading to .
3n Since liman = 0 , we cannot make a conclusion about whether diverges. n→ n=1 n!
The nth-Term Test is inconclusive for this series.
15 Example 9: Use the nth-Term Test for Divergence to determine whether the series diverges.
1
9 − 3 n=0 n
1 1 a =−9 n 3 and liman = lim 9 −3 = 9 n nn→ → n
1 Therefore, l im 0an and 9 − diverges. n→ 3 n=0 n
16 Activity Sheet with Answers
Activity: The nth-Term Test for Divergence Name:______
Determine if each of the following statements are True or False.
1. If l im 0an = , then an must converge. False n→
2. If diverges, then . False
3. If converges, then . True
4. If , then must diverge. False
5. If lim0an , then must converge. False n→
6. If , then must diverge. True
7. If diverges, then . False
8. If converges, then lim0an . False n→
Based on your results above, only statements 3 and 6 are true.
Therefore, if we can show that liman 0 , then we can conclude that a series diverges n→ using the nth-Term Test for Divergence.
th If we can show that liman = 0 , then the n -Term Test for Divergence is inconclusive. n→
17 8 9. Use a table to find l im an given an = . Round your answers to the nearest n→ n + 9 hundred thousandth.
n 1 10 100 1,000 10,000 100,000
an 0.8 0.42105 0.07339 0.00793 0.00080 0.00008
8 Based on your table, l im an = 0 . What conclusion can we make about ? n→ n + 9
th Since l im 0an = , the n -Term Test is inconclusive for this series. No conclusions can be n→ made.
( 1n )!+ 10. Use a table to find given a = . n n!
1 10 100 1,000 10,000 100,000 2 11 101 1,001 10,001 100,001
(n +1!) Based on your table, lim an = . What conclusion can we make about ? n→ n!
th (n +1!) Since lim0an , the n -Term Test can be used to conclude that diverges. n→ n!
th For each of the following, use algebraic techniques to find l im an . Then apply the n - n→ Term Test if possible.
7n 11. n=1 31n −
7n 77n an = and liman == lim 31n − nn→ → 3n − 1 3
7n Therefore, liman 0 and diverges. n→ n=1 31n −
18 −+32n 12. 2 n=4 41n −
−+32n −+32n an = and limlim0an == 41n2 − nn→→ 41n2 −
−+32n Since liman = 0 , we cannot make a conclusion about whether diverges. n→ 2 n=4 41n −
The nth-Term Test is inconclusive for this series.
n 1 13. 3 n=0 4
n n 1 1 an = 3 and limlim30an == 4 nn→→ 4
n 1 Since , we cannot make a conclusion about whether 3 diverges. n=0 4
The nth-Term Test is inconclusive for this series.
−+52nn3 14. 2 n=2 8nn+− 9 1
−+52nn3 −+52nn3 an = and limliman == − 8nn2 +− 9 1 nn→→ 891nn2 +−
−+52nn3 Therefore, lim0an and diverges. n→ 2 n=2 891nn+−
19 15. 7 n=1
a = 7 and limlim77an == n nn→→
Therefore, liman 0 and 7 diverges. n→ n=1
6 16. −+5 n=3 n
6 6 an =−+ 5 and limlim55an =−+= n nn→→ n
6 Therefore, and −+5 diverges. n=3 n
n 3n 17. (−1) n=4 56n −
n 3n n 3n an =−( 1) and liman =− lim( 1) does not exist. As we go further 56n − nn→ → 56n − 3 out in the sequence, the terms alternate between values close to and values close to 5 3 n 3n − . Therefore, and (−1) diverges. 5 n=4 56n −
20 Lesson 3: Geometric Series
Teacher Notes
Overview:
In this lesson, students will learn about geometric series. Most students have already encountered this type of series in Algebra 2. However, Algebra 2 focuses on finite geometric series, while calculus extends to infinite geometric series.
Students begin by identifying geometric series. Next, students will find the common ratio of the terms of a geometric series and use this to determine whether the series converges. If the series converges, students will calculate the sum.
I have found that many students enjoy working with this type of series since it is rather straightforward. However, students stumble when given geometric series with an initial value of the index other than n = 0. For this reason, I included examples on both the guided notes and the activity sheet that have different initial index values.
Additionally, I wanted to include a few examples of series that may not appear to be geometric at first glance. To address this, I have included series where only the numerator or denominator is raised to the nth power.
Objectives:
• Identify geometric series • Determine the convergence of geometric series • Calculate the sum of convergent geometric series
Standards:
II. Limits: LIM-7.A.3: A geometric series is a series with a constant ratio between successive terms.
II. Limits: LIM-7.A.4: If a is a real number and r is a real number such that r 1 , then a the geometric series [converges and] ar n = . n=0 1− r
21 Guided Notes with Answers
Geometric Series Name: ______
3 3 3 3 Consider the series 3+ + + + + . 2 4 8 16
What do you notice about the terms of this series?
1 The terms have a constant ratio of . 2
Series consisting of terms with a constant ratio are called geometric series. The terms of this type of series form a geometric sequence.
In general, the series given by araararararnn=++++++ 23 with a 0 and n=0 r 0 is a geometric series with common ratio r.
Example 1: Identify the ratio of the geometric series and write the series in sigma notation.
510204080+++++
The ratio is 2 and the series can be written as 52( )n . n=0
Example 2: Identify the ratio of the geometric series and write the series in sigma notation.
82111 +++++ 551040160
n 1 81 The ratio is and the series can be written as . 4 n=0 54
22 Example 3: Identify the ratio of the geometric series and write the series in sigma notation.
1392781+−++−++( ) ( )
The ratio is −3 and the series can be written as 13(− )n or (−3)n . n=0 n=0
Identifying the ratio of a geometric series is a crucial skill because the ratio determines whether the series will converge or diverge. Let’s investigate this below.
You may remember from Algebra 2 that there is a formula for the sum of the terms of a m m n 1− r finite geometric sequence: ar = a provided that r 1. n=0 1− r m m If r =1, then ar n = a(1)n = a( 1)0 + a( 1) 1 + a( 1) 2 + + a( 1)m = a( 1+ m) . n=0 n=0
We know that the value of an infinite series is the limit of its partial sums.
Using the formula above for r 1, we can represent the mth partial sum as 1− r m Sam = . Now, let’s take the limit of this partial sum. 1− r
m 1− r m 1lim− r m→ limlimSaam == mm→→ 11−−rr
If we want this sequence of partial sums to converge, we want to find what values of r will make lim r m exist. m→
If r 1 or r −1, lim r m does not exist. m→
When r 1, limr m = 0 and the sequence of partial sums converges. m→
23 So assuming r 1, we can find l im Sm . m→
m 1101−−ram 1lim− r m→ limlimSaaaam ===== mm→→ 11111−−−−−rrrrr
m n The other possibility from above was r =1 and Sarm == am(1+ ) . n=0
In this case, lim Sm = and therefore this geometric series would diverge. m→
Thus a geometric series with ratio r converges if and will have the sum a ar n = . n=0 1− r
• Notice that this characterization for the limit of the sum requires the value of the index to start at n = 0 . We will address how to calculate sums when the index starts at other values in the examples to follow.
A geometric series with ratio r diverges if r 1 .
Example 4: Determine whether the geometric series converges or diverges. If the series converges, find the sum.
n 2 3 n=0 7
2 The ratio is . 7
2 a 3 21 Since 1, the series converges to ==. 2 7 15− r 1− 7
24 Example 5: Determine whether the geometric series converges or diverges. If the series converges, find the sum.
6
n n=1 8
1 The ratio is . 8
1 Since 1, the series converges. 8
Before calculating the sum, notice that the initial value of the index is n =1. We can calculate the sum in two different ways.
Option 1: We will use the formula for the sum of a geometric series to calculate the sum of this series, but we will need to subtract a0 from the sum calculated by the formula.
66668 66 a The series converges to =−=−=−=−=aa 6. nn001 0 nn==10881877 − r 1− 8
1 Option 2: We can factor out of every term to adjust the series: 8
6666 =+++ n 23 n=1 8888 12 66161 =+++ 88888 n 61 =. n=0 88
Now, this new expression for the series has an initial index value of n = 0 and converges 6 6 a 8 6 to ===8 . 17− r 1 7 1− 8 8
25 Example 6: Determine whether the geometric series converges or diverges. If the series converges, find the sum.
n 6 − n=2 11
6 The ratio is − . 11
6 Since −1, the series converges. 11
Before calculating the sum, notice that the initial value of the index is n = 2 . We can calculate the sum in two different ways, as we did in Example 6.
Let’s choose to use the method where we subtract a0 and a1 from the sum of the geometric series whose initial index is 0.
The series converges to
nn 66 −=−−− aa01 nn==201111 a =−− aa 1− r 01 01 166 =−−−− 6 1111 1−− 11 116 =− −−1 1711 36 = . 187
26 Example 7: Determine whether the geometric series converges or diverges. If the series converges, find the sum.
1 n − (8) n=0 5
The ratio is 8 . Since 81 , the series diverges.
Example 8: Determine whether the geometric series converges or diverges. If the series converges, find the sum.
en n=1 4
The ratio is e. Since e 1, the series diverges.
27 Activity Sheet with Answers
Activity: Geometric Series Name: ______
1. Determine if each of the following is a geometric series. If it is a geometric series, write the series in summation notation.
404040 a) 12040+++++ b) −+−++++832712( ) 3927
Geometric Not geometric
n 1 120 n=0 3
c) −++−++−+2481632( ) ( ) d) +++++2345
Geometric Geometric
n −−22( )n or 12(− )n ( )n or 1( ) n=0 n=1 n=0 n=1
251256253125 e) −+++++1763215511 f) 10 +++++ 2832128
Not geometric Geometric
n 5 10 n=0 4
125 125 125 g) 1491625+++++ h) 1000+( − 250) + + − + + 2 8 32
Not geometric Geometric
n 1 1000− n=0 4
28 Determine if each of the following is a geometric series. If it is a geometric series, identify the ratio r.
n 3 2 n 2. 5− 3. n=0 3 n=0 5
2 Geometric, r =− Not geometric 3
n 19n 4. −34( ) 5. n=1 n=0 24
Geometric, r = 4 Geometric, r = 9
7 3n4 6. n 7. n n=3 13 n=2 6
1 Geometric, r = Not geometric 13
n! 51n − 8. n 9. 2 n=2 43+ n=0 842nn−+
Not geometric Not geometric
1 n n3 10. ( ) 11. n=1 3 n=0 79n −
Geometric, r = Not geometric
29 12. A geometric series diverges if r 1.
13. A geometric series converges if r 1.
a 14. For a convergent geometric series, ar n = . n=0 1− r
15. Describe two ways to calculate the sum of a convergent geometric series ar n . n=3 nn a Option 1: The series converges to araraaaaaa=−−−=−−− 012012 . nn==30 1− r
Option 2: We can factor ar3 out of every term to adjust the series:
ararararn =+++345 n=3 12 =+++ararrarr333 ( ) ( ) =(arr3 ( )n ). n=0
Now, this new expression for the series has an initial index value of n = 0 and converges 3 n ar to (arr3 =( ) ) . n=0 1− r
Students can verify algebraically that Option 1 and Option 2 give the same value for the series.
Determine whether each of the following geometric series converges or diverges. If the series converges, find the sum.
n 1 16. 4 n=0 7
1 1 r = and 1 7 7
4 14 The series converges to = . 1 1− 3 7
30 n 7 17. 5 n=0 3
7 7 r = and 1 3 3
The series diverges.
5 18. n n=1 6
1 1 r = and 1 6 6
The series converges.
555 Option 1: The series converges to =−= 1. n 1 0 n=1 661− 6
5 Option 2: We can factor out of every term to adjust the series: 6
5555 n =+++123 n=1 6666 12 55151 =+++ 66666 n 51 = . n=0 66
Now, this new expression for the series has an initial index value of n = 0 and converges 5 5 n 51 a 6 6 to ==== 1. n=0 6 61 − r 1 5 1− 6 6
31 3 n 19. − ( 11) n=0 4 r = 11 and 11 1
The series diverges.
n 5 20. −1 n=2 6
5 5 r = and 1 6 6
The series converges.
n 01 515525− Option 1: The series converges to −=−111 −− −= − . 5 n=2 6666 1− 6
2 5 Option 2: We can factor −1 out of every term to adjust the series: 6
n 234 5555 −=1111 −−−+ n=2 6666 22122 55555 = −+111 −+ −+ 66666 2 n 55 =−1. n=0 66
Now, this new expression for the series has an initial index value of n = 0 and converges 2 5 25 2 n −−1 5 5 a 6 36 25 to −1 = = = = − . n=0 6 6 1− r 5 1 6 1− 6 6
32 n 2 21. − n=0 9
2 2 r =− and −1 9 9
19 The series converges to = . 2 11 1−− 9
1 n 22. ( 7 ) n=1 8 r = 7 and 71
The series diverges.
23. Write a geometric series in summation notation that converges to a sum of 6.
Answers may vary.
n 1 Sample answer: 3 n=0 2
33 Lesson 4: p-Series
Teacher Notes
Overview:
In this lesson, students will learn about p-series. Within this lesson, students also will learn about a particular p-series, the harmonic series. Series that fit the p-series form are easy to identify and have a simple method for determining convergence. Knowing about p-series will be helpful when working with other convergence tests in later lessons.
Students begin by identifying p-series. Next, students will determine whether a given p- series will converge. Students will see examples where p is a whole number as well as a rational number. Additionally, students will need to recognize values of p when the terms of a series are presented in radical form.
In the coming lessons, students will begin a study of three tests that require knowledge of geometric series and p-series. Therefore, the summary of the activity for this lesson requires students to determine convergence of a variety of series through any method studied thus far.
Objectives:
• Identify p-series • Determine the convergence of p-series • Identify the harmonic series
Standards:
II. Limits: LIM-7.A.7: In addition to geometric series, common series of numbers include the harmonic series, the alternating harmonic series, and p-series.
34 Guided Notes with Answers p-Series Name: ______
1111 Example 1: Consider the series 1 +++++. 491625
What pattern do you notice in the terms of this series?
The denominator is the square of the term number.
What term does not seem to fit this pattern at first glance?
The first term
Can you verify that this term matches the pattern we notice?
11 Yes. Following the pattern, the first term is ==1. 112
3133 Example 2: Consider the series 3 +++++. 8964125
What pattern do you notice in the terms of this series?
The denominator is the cube of the term number.
What terms do not seem to fit this pattern at first glance?
The first term and the third term
Can we verify that these terms match the pattern we notice?
11 3 3 1 Yes. Following the pattern, the first term is ==1 and the third term is ==. 113 33 27 9
35 11111 A series of the form ppppp=++++ is a p-series where p is a positive n=1 n 1234 constant.
11111 Additionally, when p = 1, the series =++++ is called the harmonic n=1 n 1234 series.
The series in Example 1 and Example 2 are both p-series.
1111 1 We can rewrite Example 1 as 1 +++++= 2 where p = 2 . 491625 n=1 n
It is important to notice that Example 2 has a numerator other than 1. However, we know that infinite series are limits of finite sums. Based on properties of limits that we have previously studied, we know we can factor constants out of limits. Therefore, we can also factor constants out of infinite series.
kakann= where k is a constant. nn==11
3133 31 Therefore, Example 2 can be rewritten as 3 +++++= 33= 3 . 896412 5 nn==11nn
Example 3: Identify the value of p for the p-series and write the series in sigma notation.
1111 1+++++ 32243 10243125 p = 5
1 5 n=1 n
36 Example 4: Identify the value of p for the p-series and write the series in sigma notation.
1111 1+++++ 2345
1 p = 2
11 = 1/2 nn==11n n
Example 5: Identify the value of p for the p-series and write the series in sigma notation.
7777 7+++++ 44234544
1 p = 4
771 ==7 4 1/41/4 nnn===111 n nn
Example 6: Identify the value of p for the p-series and write the series in sigma notation.
11111111 11 +++++ 223344553333
4 p = 3
11 11 1 ==11 3 4/3 4/3 n=1nn n = 1nn n = 1
37 Like finding the common ratio in a geometric series, identifying the value of p in a p- series is a crucial skill because p determines whether the series will converge or diverge.
A p-series with p > 1 will converge.
A p-series with 0 < p < 1 will diverge.
Remember that p must be a positive constant.
Let’s investigate why the convergent and divergent statements above are true.
Remember that an infinite series is a sequence of partial sums. When we determine 1 whether converges, we look at the sequence of partial sums. p n=1 n
Let’s begin by looking at the partial sums of .
1 S = 1 1p
11 S =+ 2 12pp
111 S =++ 3 123ppp
1111 S =+++ 4 1234pppp
and in general,
1 1 1 1 1 Sm−1 = + + + + + 1p 2 p 3 p 4 p (m −1) p
1 1 1 1 1 1 Sm = + + + + + + . 1p 2 p 3 p 4 p(m −1) p m p
38 Think back to earlier in this course. When have we added up values of a function?
Approximating area under a curve using Riemann sums
The partial sums we have listed above look different from Riemann sums since each term is not being multiplied by x . However, if we let =x 1, we can rewrite the partial sums as follows:
1 S = (1) 1 1p
11 S =+(11) ( ) 2 12pp
111 S =++ (111) ( ) ( ) 3 123ppp
1 1 1 1 S =(1) +( 1) +( 1) + ( 1) 4 1p 2 p 3 p 4 p
and in general,
11111 Sm−1 =+++++(11111) ( ) ( ) ( ) ( ) 1234pppp (m −1) p
111111 Sm =++++++(111111) ( ) ( ) ( ) ( ) ( ) . 1234ppppp (m −1) p m
Now, we can see that each of these partial sums can be interpreted as a Riemann sum.
m 1 Let’s look at Riemann sum approximations for dx . p 1 x
A left Riemann sum approximation with =x 1 would be:
m 1 11111 dx 11111++++ + . p pppp( ) ( ) ( ) ( ) p ( ) 1 x 1234 (m −1)
This would correspond to the partial sum Sm−1 . 39 1 We know that p 0 and thus fx()= is a decreasing function on the interval (0,) . x p
Left Riemann sum approximation [4].
Therefore, the left Riemann sum approximation is an overestimate of the integral.
m 1 That is, Sdx . m−1 p 1 x
For 01p ,
m 1 limlimSdxm−1 mm→→ p 1 x m x(−+p 1) = lim m→ −+p 1 1 m(−pp +−11) + 1( ) =−lim. m→ −+−+pp11
When , −+p 1 is positive and therefore m(−+p 1) grows without bound as m(−pp +−11) + 1( ) m →. Thus lim − is infinite. This implies that lim Sm−1 is infinite and m→ m→ −+−+pp11 1 the sequence of partial sums diverges. Thus, diverges. p n=1 n
40 For p =1,
m 1 l im lS im dxm−1 mm→→ 1 1 x m = l im l n x m→ 1 =−l im l n l n 1m m→ ( ) =−l im l n 0 m m→ ( ) =.
Thus l im Sm−1 is infinite and therefore the sequence of partial sums diverges. Thus, m→ 1 diverges. p n=1 n
A right Riemann sum approximation with =x 1 would be:
m 1 111 11 dx 1111++++ +1 . p ppp( ) ( ) ( ) p ( ) p ( ) 1 x 234 (m −1) m
1 This would correspond to the partial sum S − (1) . m 1p
1 We know that p 0 and thus fx()= is a decreasing function on the interval (0,) . x p
Right Riemann sum approximation [4].
41 Therefore, the right Riemann sum approximation is an underestimate of the integral.
11m 11m That is, S−1 dx , so S+1 dx . m pp( ) m pp( ) 1 1 x 1 1 x
For p 1,
11m limlim1Sdxm +( ) mm→→ pp 1 1 x 11m =+(1lim) dx ppm→ 1 1 x m 1 x(−+p 1) =+(1lim) 11p m→ −+p 1 11m(−pp +−11) + ( ) =+− (1lim.) p m→ 111 −+−+pp
11m(−pp +−11) + ( ) Thus, lim1lim.S +− m p ( ) mm→→ 111 −+−+pp
When p 1, −+p 1 is negative and therefore m(−+p 1) converges to 0 as m →. Thus 11m(−pp +11) ( − + ) (1) +− lim exists. This implies that l im Sm converges. Since the limit p m→ 1m→ −pp + 1 − + 1 1 of partial sums converges, converges. p n=1 n
Therefore, converges when p 1 and diverges when 01p .
We will usually assign p-series an initial index value of n =1, however the initial value of the index can be any finite positive value. This will not affect the convergence of a p- series.
42 Using this information, we can determine whether the series in Examples 3 through 6 converge.
Example 3: p = 5 and 51 The series converges.
1 1 Example 4: p = and 01 The series diverges. 2 2
1 1 Example 5: p = and 01 The series diverges. 4 4
4 4 Example 6: p = and 1 The series converges. 3 3
Example 7: Determine whether the p-series converges or diverges.
11111 6/76/76/76/76/7=+++++ 1 n=1 n 2345
6 6 p = and 01 7 7
The series diverges.
Example 8: Determine whether the p-series converges or diverges.
99999 =+++++ 9 n=1 n 2345 p = and 1
The series converges.
43 Example 9: Determine whether the p-series converges or diverges.
3 3 3 3 3 = 3 + + + + + 666 6 6 n=1 n 2345
1 1 p = and 01 6 6
The series diverges.
Example 10: Determine whether the p-series converges or diverges.
8.38.38.38.38.3 =+++++ 8.3 n=1 n 2345 p =1 and 0 1 1
The series diverges. This is a multiple of the harmonic series.
44 Activity Sheet with Answers
Activity: p-Series Name: ______
1. Determine if each of the following is a p-series. If it is, write the p-series in summation notation.
1111 7777 a) 1+++++ b) 7 +++++ 2345−−−−1111 2345
Not a p-series. p-series with p =1; multiple of the
harmonic series
7 n=1 n
1234 17171717 5555 c) 17 +++++ d) 77777+++++ 223344559999 2222
10 p-series with p = Not a p-series. 9
1717 = 9 10/9 nn==11nn n
−−−−−55555 1414 e) 12345+++++ f) 4 +++++ 22716125 p-series with p = 5 p-series with p = 3 1 4 −5 n = 5 3 nn==11n n=1 n
45 Identify p in each of the following p-series. Then, determine if the series converges or diverges.
1 1 2. 3/7 3. e n=1 n n=1 n
3 3 p = and 01 pe= and e 1 7 7
The series diverges. The series converges.
1 4. Write a p-series in summation notation with p = and whose first term is 9. 45
9 1 /45 n=1 n
5. Write a p-series in summation notation with p = 6 and whose first term is 1.
1 6 n=1 n
Determine whether each of the following series converges or diverges. Use any method we have studied.
83+ n n 6. 7. 53( ) n=1 9n n=2
Use the nth-Term Test for Divergence. Geometric series
83+ n 831+ n an = and limliman == r = 3 and 31 9n nn→→ 93n
Therefore, liman 0 and the series diverges. The series diverges. n→
46 13 −2n 8. 9. 2 5 n=1 nn n=0 3 p-Series Geometric series
11 11 p = and 1 r = 2 and 21 5 5
The series converges. The series diverges.
5.2n −3 10. 3 11. 8 n=2 n n=1 n
Use the nth-Term Test for Divergence. p-Series
5.2n 5.2n an = and limliman == p = 8 and 81 n3 nn→→ n3
Therefore, liman 0 and the series diverges. The series converges. n→
n 1 4 12. 13. n=0 6 n=1 n
Geometric series p-Series
1 1 1 1 r = and 1 p = and 01 6 6 2 2
The series converges. The series diverges.
47 4n 16.2 14. 15. n n=3 n=0 35
Use the nth-Term Test for Divergence. Geometric series
4 4 an = 16.2 and limlim16.216.2an == r = and 1 nn→→ 5 5
Therefore, l im 0an and the series diverges. The series converges. n→
16. Complete the following chart.
General form of Series diverges if … Series converges if… series
th The n -Term Test for an Not applicable Divergence nk=
Geometric Series ar n r 1 r 1 nk=
1
p-Series p 01p p 1 nm= n
Note: Let k represent any integer and let m represent any positive integer.
17. The nth-Term Test for Divergence can only determine that a series diverges.
18. We can calculate the sum of a convergent geometric series with an initial index value a of n = 0 using the formula . 1− r
1 19. In a p-Series in the form p , p must be a positive constant. nm= n
48 Lesson 5: Direct Comparison Test
Teacher Notes
Overview:
In this lesson, students will learn about the Direct Comparison Test. To apply this test, students must use prior knowledge of geometric series and p-series. Students should be familiar with characteristics of each type of series and be able to identify when these series converge or diverge. Additionally, when given two series, students will need to determine which series has larger or smaller terms.
In the activity, students will identify a series that is similar to a given series. Next, students will describe scenarios that are supported by the Direct Comparison Test. I have also included a question asking students to describe the two cases when the Direct Comparison Test is inconclusive. Additionally, students will need to apply the Direct Comparison Test to a given series.
As the final activity, I have asked students to decide what test or method would be best to use to determine if a given series converges or diverges.
Objectives:
• Identify series similar to a provided series • Apply convergence rules of geometric series and p-series • Given two series, justify which series has smaller terms • Given two series, justify which series has larger terms • Apply the Direct Comparison Test
Standards:
II. Limits: LIM-7.A.8: The comparison test is a method to determine whether a series converges or diverges.
49 Guided Notes with Answers
Direct Comparison Test Name: ______
1 Example 1: We are given the series and asked to determine whether the series 2 n=1 n +1 converges or diverges. What methods might we try in order to answer this question?
1. The nth-Term Test for Divergence 1 1 a = and limlim0an == n n2 +1 nn→→ n2 +1
1 Since l im 0an = , we cannot make a conclusion about whether diverges. n→ 2 n=1 n +1 The nth-Term Test is inconclusive for this series.
2. Geometric Series 1 n The series 2 is not of the form ar . Additionally, we cannot factor out n=1 n +1 n=0 a value to adjust the given series to fit the form. This is not a geometric series. We cannot use what we know about geometric series to determine whether the series converges or diverges.
3. p-Series 1 1 The series 2 does not fit the form p . This is not a p-series. We cannot n=1 n +1 n=1 n use what we know about p-series to determine whether the series converges or diverges.
None of the methods we have studied so far is applicable to this particular series.
What if instead of matching a known series type exactly, we looked for a type that was similar to the series we are working with?
50 1 The series is similar to a p-series. 2 n=1 n +1
What p-series would we say is most similar to ?
1
2 n=1 n
What do we know about the series we wrote above?
In the series , p = 2 and 21 . Therefore, converges.
Let’s list a few terms of the original series we were given and the new series we have compared it to.
11111 Original: 2 =++++ n=1 n +1251017
1111 p-Series: 2 =++++1 n=1 n 4916
Let’s compare the two series term by term.
1 11 11 11 1 2 54 10 9 17 16
We can see that each term of the original series is less than the corresponding term of the series we selected.
Additionally, we can compare the expression for the nth term for each series.
1 1 a = and b = n n2 +1 n n2
51 What do we notice about these expressions?
The numerator is the same for each expression. However, an has a larger denominator and will result in smaller terms.
1 Since we know that the series converges, what do you think this means about our 2 n=1 n original series?
Since the terms of the original series are positive and smaller than the terms of the p- series, the partial sums of the original series must be positive and smaller than the partial sums of the p-series. We know that the sequence of partial sums of the p-series converges, and thus the p-series converges.
Therefore, it makes sense that the sequence of partial sums of the original series must converge as well, resulting in the original series converging.
Although this is not a formal proof, there is a proof which you may learn in a college calculus course.
This leads us to the next convergence test we will study: the Direct Comparison Test.
Direct Comparison Test
Given two series an and bn , let 0 abnn for all n. n=1 n=1
1. If bn converges, then an converges. n=1 n=1
2. If diverges, then diverges.
To use this test, we must find a series similar to the series we are given. We should choose a series that is either a geometric series or a p-series since we can easily determine whether these series converge or diverge.
52 Additionally, notice that this test requires both an and bn to consist of positive n=1 n=1 terms. This is an important condition to verify before applying the Direct Comparison Test.
Let’s investigate why this condition is important. We will assume we only need abnn .
−1 1 a = Let n and bn = 2 . nn==11n nn==11n
It is true that for all terms since every an is negative and every bn is positive. We −1 1 know that diverges since it is the negative of the harmonic series and is a 2 n=1 n n=1 n p-series that converges. So even with , when the terms of the series are not all
positive, we cannot use the convergence of bn to determine the convergence of an . n=1 n=1
Also even with , we cannot use divergence of to determine the divergence of .
Therefore, we must be sure that we are working with series consisting of positive terms when we apply this test.
53 Example 2: Use the Direct Comparison Test to determine whether the series converges or diverges.
5
3 n=1 n
11 We will compare this series to the p-series = . 3 1/3 nn==11n n
5 1 Note that both the original series and new series consist of positive terms. 3 1 /3 n=1 n n=1 n
1 1 1 In the series , p = and 01. Therefore, diverges. 1 /3 n=1 n 3 3
1 5 Let a = and b = . n 3 n n 3 n
Notice that an and bn have the same denominator. However, has a larger numerator ab and therefore nn. Since we know that an diverges, we know that bn diverges n=1 n=1 as well by the Direct Comparison Test.
Thus, the series diverges.
We could have also determined that the original series diverges by noticing that it is a multiple of the p-series we used in the comparison.
54 Example 3: Use the Direct Comparison Test to determine whether the series converges or diverges.
2n n n=1 35+
22n n We will compare this series to the geometric series . n = nn==1133
n n 2 2 Note that both the original series n and new series consist of positive n=1 35+ n=1 3 terms.
2 2 In the series , r = and 1. Therefore, converges. 3 3
2n 2n Let a = and b = . n 35n + n 3n
Notice that an and bn have the same numerator. However, has a larger denominator and therefore abnn . Since we know that bn converges, we know that an n=1 n=1 converges as well by the Direct Comparison Test.
Thus, the series converges.
55 Example 4: Use the Direct Comparison Test to determine whether the series converges or diverges.
8 n=1 n + 3
8 We will compare this series to the harmonic series . n=1 n
8 Note that both the original series and new series consist of positive terms. n=1 n + 3
In the series , p =1 and 0 1 1 . Therefore, diverges.
8 8 Let a = and b = . n n + 3 n n
Notice that an and bn have the same numerator. However, an has a larger denominator and therefore abnn .
We know that bn diverges, but we cannot make a conclusion about an using the n=1 n=1 Direct Comparison Test.
We will investigate another method to determine the convergence or divergence of a series in the next lesson that will address this situation.
56 Example 5: Use the Direct Comparison Test to determine whether the series converges or diverges.
1
5 n=1 n +1
11 We will compare this series to the p-series = . 5 5/2 nn==11n n
1 1 Note that both the original series and new series consist of positive 5 5/2 n=1 n +1 n=1 n terms.
5 5 In the series , p = and 1. Therefore, converges. 2 2
1 1 Let an = and bn = . n5 +1 n5
Notice that an and bn have the same numerator. However, has a larger denominator ab and therefore nn. Since we know that bn converges, we know that an n=1 n=1 converges as well by the Direct Comparison Test.
Thus, the series converges.
57 Example 6: Use the Direct Comparison Test to determine whether the series converges or diverges.
1
4 n=1 72n −
1 We will compare this series to a multiple of a p-series with p = . We will use the series 4 11 = . 4 1/4 nn==117 n 7n
1 1 Note that both the original series and new series consist of positive 4 1 /4 n=1 72n − n=1 7n terms.
1 In the series , and 01. Therefore, diverges. 4
1 1 Let a = and b = . n 7 4 n n 724 n −
Notice that an and bn have the same numerator. However, has a smaller denominator ab and therefore nn. Since we know that an diverges, we know that bn diverges n=1 n=1 as well by the Direct Comparison Test.
Thus, the series diverges.
58 Activity Sheet with Answers
Activity: Direct Comparison Test Name: ______
Determine if each of the following is similar to a geometric series or a p-series. Then provide the series you would use with the Direct Comparison Test.
52n + 1 1. 2. n 3 n=1 4 n=1 31n +
Geometric series p-series
n 5 1 1 1 = 33 n=1 4 nn==1133nn
1 6n 3. 4. 7 5 n n=1 109 n − n=1 117+ p-series Geometric series
111 6 n = 7755 nn==1110 nn10 n=1 11
Use the words “smaller” and “larger” to complete the following statements describing the results of the Direct Comparison Test.
5. If the larger series converges, the smaller series must also converge.
6. If the smaller series diverges, the larger series must also diverge.
59 Determine whether each of the following series converges or diverges using the Direct Comparison Test.
2 7. 6 n=1 n +12
22 We will compare this series to the p-series = . 6 3 nn==11n n
2 2 Note that both the original series and new series consist of positive 6 3 n=1 n +12 n=1 n terms.
2 In the series , and . Therefore, converges. p = 3 31 3 n=1 n
2 2 Let an = and bn = . n6 +12 n6
Notice that an and bn have the same numerator. However, has a larger denominator ab and therefore nn. Since we know that bn converges, we know that an n=1 n=1 converges as well by the Direct Comparison Test.
Thus, the series converges.
60 8n 8. n n=1 74−
88n n We will compare this series to the geometric series . n = nn==1177
n n 8 8 Note that both the original series n and new series consist of positive n=1 74− n=1 7 terms.
8 8 In the series , r = and 1. Therefore, diverges. 7 7
8n 8n Let a = and b = . n 7n n 74n −
Notice that an and bn have the same numerator. However, has a smaller denominator ab and therefore nn. Since we know that an diverges, we know that bn diverges n=1 n=1 as well by the Direct Comparison Test.
Thus, the series diverges.
61 0 ab 9. Given two series an and bn with nn, describe two scenarios that are n=1 n=1 inconclusive when using the Direct Comparison Test.
Scenario 1: We can prove that an converges. Then we cannot make a conclusion n=1 about bn using the Direct Comparison Test. n=1
Scenario 2: We can prove that diverges. Then we cannot make a conclusion about
using the Direct Comparison Test.
State the method you would use to determine the convergence of each of the following series. You do not need to determine whether the series converges or diverges.
6 9 10. 11. 5 3 n n=1 n n=1 5 p-series Geometric series
2n 11n − 8 12. n 13. n=1 37+ n=1 57n +
Direct Comparison Test The nth-Term Test for Divergence
62 Lesson 6: Limit Comparison Test
Teacher Notes
Overview:
In this lesson, students will learn about the Limit Comparison Test. Similar to the Direct Comparison Test, students will need to use prior knowledge of geometric series and p- series. When given a series that does not fit the form of a geometric series or a p-series, students will identify a similar series that is one of these types of series. Students will then apply the Limit Comparison Test to make a conclusion about the original series.
This test is important since it often can be used in place of the Direct Comparison Test, and can be applied in more situations. Unlike the Direct Comparison Test, students do not need to determine what series produces larger or smaller terms. Instead, students must take a limit. Students are very familiar with limits since the beginning of the calculus course focuses on computing limits through various methods.
The activity for this lesson is very straightforward. Students are asked to apply the Limit Comparison Test to a variety of series. Additionally, students are asked to identify when they would attempt to use this test. Finally, students will describe what limits will allow them to apply the Limit Comparison Test.
Objectives:
• Identify series similar to a provided series • Apply convergence rules of geometric series and p-series • Find the limit of a ratio • Apply the Limit Comparison Test
Standards:
II. Limits: LIM-7.A.9: The limit comparison test is a method to determine whether a series converges or diverges.
63 Guided Notes with Answers
Limit Comparison Test Name: ______
7 Example 1: We are given the series and asked to determine whether the series n n=1 65− converges or diverges. What methods might we try in order to answer this question?
Note: Our instincts should lead us to the Direct Comparison Test, but let’s investigate the other methods we have learned as well.
1. The nth-Term Test for Divergence 7 7 an = and limlim0an == 65n − nn→→ 65n −
7 Since lim0an = , we cannot make a conclusion about whether diverges. n→ n n=1 65− The nth-Term Test is inconclusive for this series.
2. Geometric Series The series is not of the form ar n . Additionally, we cannot factor out n=0 a value to adjust the given series to fit the form. This is not a geometric series. We cannot use what we know about geometric series to determine whether the series converges or diverges.
3. p-Series 1 The series does not fit the form p . This is not a p-series. We n=1 n cannot use what we know about p-series to determine whether the series converges or diverges.
4. Direct Comparison Test 71n The series is similar to the geometric series . n = 7 nn==1166
64 n 7 1 Note that both the original series n and new series 7 consist of n=1 65− n=1 6 positive terms.
n 1 1 1 In the series 7, r = and 1. Therefore, converges. n=1 6 6 6
7 7 Let a = and b = . n 65n − n 6n
Notice that an and bn have the same numerator. However, has a smaller
denominator and therefore abnn .
Notice that this is a different scenario than we need in order to apply the Direct Comparison Test.
Although we know that bn converges, we cannot make a conclusion about n=1 an using the Direct Comparison Test. n=1
None of the methods we have studied so far is applicable to this particular series.
However, there is another test we can use to determine the convergence or divergence of a series: The Limit Comparison Test.
As in the Direct Comparison Test, the Limit Comparison Test will require us to use a series that is similar to the series we are given.
65 Limit Comparison Test
an Suppose that an 0 , bn 0 , and l im = L where L is finite and positive. Then the n→ bn two series an and bn either both converge or both diverge.
Let’s take a look at the possibilities for L to see why L must be finite and positive in order to apply the Limit Comparison Test.
Since we must begin with and , we know that the limit of the ratio of these terms must be nonnegative if the limit exists. Therefore, we know L 0 .
Case 1: If L is a positive finite value, we can conclude that the limit of the ratio of terms a of n is this same positive finite value. Therefore when we look far enough out in the bn series, the terms of are about L times the size of the terms of bn .
So the partial sums for converge if and only if the partial sums for converge. Likewise, the partial sums for diverge if and only if the partial sums for diverge.
Case 2: If L = 0 or the limit does not exist, the Limit Comparison Test is inconclusive. The following examples demonstrates this.
1 1 Let’s use the series 2 and . Both series consist of positive terms. n=1 n +1 n=1 n
1 1 Let a = and b = . n n2 +1 n n
66 a Now, find l im n . n→ bn
a 111 nn limlimlimlim0n ==== nnnn→→→→ 222 bnnnnn +++1111
Let’s change the labels and find once again.
1 1 Let a = and b = . n n n n2 +1
a 1 1 1nn22++ 1 1 limn = lim = lim = lim = n→ n →2 n → n → bn n n+11 n n
1 We know 2 converges by the Direct Comparison Test (see Lesson 5) and we n=1 n +1 1 know that is the harmonic series which diverges. So in these examples, when n=1 n L = 0 or the limit does not exist, one series converges while the other diverges.
Note: There are other examples where both series converge or both series diverge even when L is not finite and positive.
To summarize, we have shown that we can only apply the Limit Comparison Test to two series consisting of positive terms if the limit of the ratio of their terms is a finite, positive value.
67 7 1 n Going back to Example 1, we can compare to . When we were trying n n=1 65− n=1 6 n 1 to apply the Direct Comparison Test, we used 7as the comparison series but n=1 6 n 1 were unable to apply the test because 7 has smaller terms than . n=1 6
Since we do not need to decide what series has larger or smaller terms in order to use the Limit Comparison Test, we will use the simpler version of the geometric series in the comparison.
n n 1 1 1 1 The series is a geometric series with r = and 1. Therefore, n=1 6 6 6 n=1 6 converges.
7 1 n Note that both the original series and new series consist of positive n n=1 65− n=1 6 terms.
Let’s apply the Limit Comparison Test.
n 7 1 Let an = n and bn = . 65− 6
a Now, find lim n . n→ bn
7 n a n 76n 76( ) limn = lim65− = lim = lim = 7 n→ n →n n →nn n → bn 1 6−− 5 1 6 5 6
Since L = 7 and 7 is finite and positive, both series converge.
68 What if we interchanged the labels before calculating the limit?
n 1 7 Let an = and bn = n . 6 65−
a Find l im n . n→ bn
n 1 nn an 6 165651−− limlimlimlim==== nnnn→→→→ 7 n n bn 677 76( ) 65n −
1 1 Since L = and is finite and positive, both series converge. 7 7
Notice that it did not matter which series we labeled as an or bn . Both limits produced positive, finite values.
When using the Limit Comparison Test, we can choose to label the given series as either or . In some cases, we may find that changing how we label the series may help us simplify the limit.
However, to be consistent in these notes, we will label the given series as .
69 Example 2: Use the Limit Comparison Test to determine whether the series converges or diverges.
1 n=1 n + 4
11 We will compare this series to the p-series . = 1/2 nn==11n n
1 1 Note that both the original series and new series 1 /2 consist of positive n=1 n + 4 n=1 n terms.
1 1 In the series , p = and 01. Therefore, diverges. 2 2
1 1 Let a = and b = . n n + 4 n n
a 111 nn limlimlimlim1n ==== nnnn→→→→ bn nnnn+++444 1
Since L =1 and 1 is finite and positive, both series diverge.
70 Example 3: Use the Limit Comparison Test to determine whether the series converges or diverges.
3nn 5 12 +− 73 n=1 4 8nn 9++
n2 1 = We will compare this series to the p-series 75. nn==11nn
3nn 5 12 +− 1 Note that both the original series 73 and new series 5 consist of n=1 4 8nn 9++ n=1 n positive terms.
In the series , p = 5 and 51 . Therefore, converges.
351nn2 +− 1 Let a = and b = . n 489nn73++ n n5
a 3511351353nnnnnnnn225765+−+−+− limlimlimlimn ==== nnnn→→→→ 7357373 bnnnnnnnn 489489++++++ 14894
3 3 Since L = and is finite and positive, both series converge. 4 4
71 Example 4: Use the Limit Comparison Test to determine whether the series converges or diverges.
5n n n=0 27+
55n n We will compare this series to the geometric series . n = nn==0022
n n 5 5 Note that both the original series n and new series consist of positive n=0 27+ n=0 2 terms.
n 5 5 5 In the series , r = and 1. Therefore, diverges. n=0 2 2 2
n n 5 5 Let an = n and bn = . 27+ 2
5n a n 522nnn limlimlimlim1n ====27+ nnnn→→→→ n nnn bn 5 27++ 527 2
Since L =1 and 1 is finite and positive, both series diverge.
72 Example 5: Use the Limit Comparison Test to determine whether the series converges or diverges.
3n n n=1 41−
33n n We will compare this series to the geometric series . n = nn==1144
n n 3 3 Note that both the original series n and new series consist of positive n=1 41− n=1 4 terms.
3 3 In the series , r = and 1. Therefore, converges. 4 4
n n 3 3 Let an = n and bn = . 41− 4
3n a n 344nnn limlimlimlim1n ====41− nnnn→→→→ n nnn bn 3 41−− 341 4
Since L =1 and 1 is finite and positive, both series converge.
73 Activity Sheet with Answers
Activity: Limit Comparison Test Name: ______
a 1. The Limit Comparison Test requires us to calculate l im n = L . What must be true n→ bn about L in order to be able to apply this test?
L must be positive and finite.
Determine whether each of the following series converge or diverge using the Limit Comparison Test.
5 2. 4 3 n=1 n + 8
11 We will compare this series to the p-series = . 4 3 3/4 nn==11n n
5 1 Note that both the original series and new series consist of positive 4 3 3/4 n=1 n + 8 n=1 n terms.
3 3 In the series , p = and 01. Therefore, diverges. 4 4
5 1 Let an = and bn = . 4 n3 + 8 n3/4
a 5 1 544nn33 5 limn = lim = lim = lim = 5 n→ n →43 4 3 n → 4 3 n → 4 3 bn n+8 n n + 81 n + 8
Since L = 5 and 5 is finite and positive, both series diverge.
74 84n + 3. n n=0 1 3 7+
88n n We will compare this series to the geometric series . n = nn==001313
n n 84+ 8 Note that both the original series n and new series consist of positive n=0 1 3 7+ n=0 13 terms.
8 8 In the series , r = and 1. Therefore, converges. 13 13
n n 84+ 8 Let an = n and bn = . 13 7+ 13
84n + nn a n 84nn+ 13 13( 84+ ) limlimlimlim1n ====137+ nnnn→→→→ n nn nn bn 8 137+ 8 8( 137+ ) 13
Since L =1 and 1 is finite and positive, both series converge.
75 8nn 3 72 +− 4. 3 n=1 92nn−
n2 1 = We will compare this series to the harmonic series 3 . nn==11nn
8nn 3 72 +− 1 Note that both the original series 3 and new series consist of positive n=1 92nn− n=1 n terms.
We know that the harmonic series diverges.
8nn 3 72 +− 1 Let a = and b = . n 92nn3 − n n
a 837nnnnnnnn2232+−+−+− 18378378 limlimlimlimn = = == nnnn→→→→ 333 bnnnnnnnn 92921929−−−
8 8 Since L = and is finite and positive, both series diverge. 9 9
5. When asked to determine if a series converges or diverges, what would lead you to try the Limit Comparison Test rather than other methods?
Sample Answer: If the series resembled a geometric series or p-series, but did not exactly match the form of either series type, I would try Limit Comparison Test. I could also try to apply the Direct Comparison Test, but then I would need to determine which series had larger or smaller terms. With the Limit Comparison Test, I do not need to know which is a larger or smaller. I can find lim n and if the limit is positive and finite, I can make a n→ bn conclusion about the given series.
76 Let’s assume an 0 and bn 0 . You have done the work to calculate the following limits. If you arrived at each of the following conclusions, would you be able to apply the Limit Comparison Test? Why or why not?
a 6. l im n = n→ bn
We cannot apply the Limit Comparison Test. The limit is infinite. In the Limit Comparison Test, the limit must be finite.
a 1 7. l im n = n→ bn 3
1 We can apply the Limit Comparison Test. The limit is . This is positive and finite. 3
a 8. lim1 n = n→ bn
We can apply the Limit Comparison Test. The limit is 1. This is positive and finite.
a 9. limn = 0 n→ bn
We cannot apply the Limit Comparison Test. The limit is 0. This is not a positive value.
a 10. limn = 7 n→ bn
We can apply the Limit Comparison Test. The limit is 7 . This is positive and finite.
77 Lesson 7: Ratio Test
Teacher Notes
Overview:
In this lesson, students will learn about the Ratio Test. To apply this test, students will take the limit of the ratio of two terms from the same series. This test does not require students to apply information they’ve previously learned regarding geometric series or p- series. The Ratio Test is useful when working with series involving factorials, as well as series that combine different types of functions.
The Ratio Test will be extremely useful for students who continue to study calculus. For example, they will use this test to discover the radius of convergence for Taylor and Maclaurin polynomial approximations.
The activity for this lesson requires students to apply the Ratio Test to several series as well as describe the conditions of the Ratio Test. Additionally, students are asked to simplify an expression involving factorials. As the final activity, students are given several different series. For each series, they must identify an appropriate convergence/divergence test.
Objectives:
• Find the limit of a ratio • Apply the Ratio Test
Standards:
II. Limits: LIM-7.A.11: The ratio test is a method to determine whether a series of numbers converges or diverges.
78 Guided Notes with Answers
Ratio Test Name: ______
n Example 1: We are given the series and asked to determine whether the series n n=1 5 converges or diverges. What methods might we try in order to answer this question?
1. The nth-Term Test for Divergence n n an = and limlim0an == by the Law of Dominance 5n nn→→ 5n
n Since l im 0an = , we cannot make a conclusion about whether diverges. n→ n n=1 5 The nth-Term Test is inconclusive for this series.
2. Geometric Series The series is not of the form ar n . Additionally, we cannot factor out a n=0 value to adjust the given series to fit the form. This is not a geometric series. We cannot use what we know about geometric series to determine whether the series converges or diverges.
3. p-Series 1 The series does not fit the form . This is not a p-series. We cannot p n=1 n use what we know about p-series to determine whether the series converges or diverges.
4. Direct Comparison Test and Limit Comparison Test
The series is not similar to a geometric series or a p-series. We do not have
a known series to use in a comparison. We cannot make a conclusion about this series using the Direct Comparison Test or the Limit Comparison Test.
79 None of the methods we have studied so far is applicable to this particular series.
However, there is another test we can use to determine the convergence or divergence of a series: The Ratio Test.
The Ratio Test is useful since we do not need to compare the given series to another series. We only need to use the given series and the expression for its terms. We can also use the Ratio Test for series with negative terms.
Ratio Test
Let an be a series with nonzero terms.
a 1. The series converges if l im 1 n+1 . n→ an a a 2. The series diverges if lim1 n+1 or lim n+1 =. n→ n→ an an a a 3. The Ratio Test is inconclusive if lim1 n+1 = or lim n+1 does not exist. n→ n→ an an
The absolute values are important, especially for series with some positive terms and some negative terms, but where the positive and negative values do not alternate from one term to the next. In a case such as this, the limit of the ratio of successive terms with the absolute value would exist, but the limit without the absolute value would not exist. We will not investigate a case such as this in the notes but interested students can ask to see an example later.
Note: For interested students, explore the following series:
n sin + 42 22222 =+ + −+ −+ + n n=0 22481664 1 1111 = 2.+ + − + −+ + 2 4816 64
80 a The Ratio Test requires us to find l im n+1 . Unlike the Limit Comparison Test, we n→ an cannot calculate the limit of the reciprocal and arrive at the same conclusion. For a a 1 example, if l im 5 n+1 = , then l im n = . This would lead to two different n→ n→ an an+1 5 conclusions using the Ratio Test.
n Let’s apply the Ratio Test to the series in Example 1: . n n=1 5
n Note that has nonzero terms. n n=1 5
n Let a = . n 5n
a nn+1 limlimn+1 = nn→→ nn+1 an 55 n +15n =lim n→ 5n+1 n 51n n + =lim n→ 5n+1 n 51n n + =lim n n→ 55 n 11n + =lim n→ 5 n 1 = 5
a 1 1 Since lim n+1 = and 1, the series converges by the Ratio Test. n→ an 5 5
81 Example 2: Use the Ratio Test to determine whether the series converges or diverges.
n 2 n n=1 3
n 2 Note that n has nonzero terms. n=1 3
n 2 Let ann = . 3
n+1 2 (n + 1) a 3 limn+1 = lim nn→ → n an 2 n 3 (nn+1)( 2)nn+1 ( 2) =lim n→ (33)nn+1 ( ) (n +1)( 2)nn+1 ( 3) =lim n→ (32)nn+1 n( ) 2nn+1 3n + 1 =lim n→ 23nn+1 n 2+ 2nn 3n 1 =lim n→ 2nn 3 3 n 2 1n + 1 =lim n→ 13 n 2 = 3
a 2 2 Since lim n+1 = and 1, the series converges by the Ratio Test. n→ an 3 3
82 Example 3: Use the Ratio Test to determine whether the series converges or diverges.
8n
6 n=2 n
8n Note that has nonzero terms. 6 n=2 n
8n Let a = . n n6
a 88nn+1 limlimn+1 = nn→→ 6 6 ann (n +1) 8n+16n =lim n→ (n +1)6 8n 8n+16n =lim n→ 8n (n +1)6
n 6 88 n =lim n n→ 81n +
6 8 n =lim n→ 11n + = 8
a Since lim8 n+1 = and 81 , the series diverges by the Ratio Test. n→ an
83 Example 4: Use the Ratio Test to determine whether the series converges or diverges.
6 n=1 n
6 Note that has nonzero terms. n=1 n
6 Let a = . n n
a 66 limn+1 = lim nn→ → an n+1 n 6 n =lim n→ n +16 6n = lim n→ 61(n + ) n = lim n→ n +1 =1
a Since lim1 n+1 = , the Ratio Test is inconclusive for the series . n→ an
However, we can identify as a multiple of the harmonic series. Therefore, we can conclude that diverges by other means.
84 Example 5: Use the Ratio Test to determine whether the series converges or diverges.
3
4 n=2 n +1
3 Note that has nonzero terms. 4 n=2 n +1
3 Let a = . n n4 +1
a 33 limlimn+1 = nn→→ 4 4 ann (n ++11) +1 31n4 + =lim n→ (n ++11)4 3 n4 +1 = lim n→ (n ++11)4 =1
a Since lim1 n+1 = , the Ratio Test is inconclusive for the series . n→ an
3 However, we can compare to the series . Both series consist of positive 4 n=2 n terms, and the terms of are less than the terms of . Since we know is a convergent p-series, we can conclude that converges as well by the Direct
Comparison Test.
85 Example 6: Use the Ratio Test to determine whether the series converges or diverges.
9n n=2 n!
9n Note that has nonzero terms. n=2 n!
9n Let a = . n n!
a 99nn+1 limn+1 = lim nn→ → an ( n+1) ! n ! 9!n+1 n =lim n→ (n +1) ! 9n 9!n+1 n =lim n→ 9n (n + 1) ! 9 9n n ! =lim n→ 9n (nn+ 1) ! 91 =lim n→ 11(n + ) = 0
a Since lim0 n+1 = and 01 , the series converges by the Ratio Test. n→ an
86 Example 7: Use the Ratio Test to determine whether the series converges or diverges.
(n −1!)
7 n=1 n
(n −1!) Note that has nonzero terms. 7 n=1 n
(n −1!) Let a = . n n7
a (nn+−−11 !1) ! ( ) limlimn+1 = nn→→ 7 7 ann (n +1) nn! 7 =lim n→ (n +1)7 (n −1!) nn! 7 =lim n→ (n −1!) (n +1)7
7 nn−( 1!) n =lim n→ (nn−+1) !1
7 nn =lim n→ 11n + =
a Since lim n+1 =, the series diverges by the Ratio Test. n→ an
87 Example 8: Use the Ratio Test to determine whether the series converges or diverges.
n −9 n n=1 7
n −9 Note that n has nonzero terms. n=1 7
n −9 Let ann = . 7
nn+1 a (nn+−−199)( ) ( ) limlimn+1 = nn→→ nn+1 an 77
n+1 (n +−19)( ) 7n =lim n→ 7n+1 n(−9)n
n+1 (−9) 71n n + =lim n→ (−9)n 7n+1 n
n (−99) −( ) 71n n + =lim n→ (−9)n 77 n n −+911 n =lim n→ 17 n 9 = 7
a 9 9 Since lim n+1 = and 1, the series diverges by the Ratio Test. n→ an 7 7
88 Activity Sheet with Answers
Activity: Ratio Test Name: ______
Complete the following statements regarding the Ratio Test.
1. The series an must have nonzero terms.
a 2. If l im 1 n+1 , then converges. n→ an
a a 3. If l im 1 n+1 or l im n+1 =, then diverges. n→ n→ an an
a 4. If lim1 n+1 = , the Ratio Test is inconclusive. n→ an
5. Describe an advantage of using the Ratio Test instead of the Direct or Limit Comparison Tests.
With the Ratio Test, we do not need to use an additional series to make a conclusion about the given series. Both the Direct Comparison Test and Limit Comparison Test require us to identify a series that is similar to the given series.
n! 6. Simplify the following factorial expression: . (n − 5!)
n! n( n −1) ( n − 2) ( n − 3) ( n − 4) ( n − 5) ! = (nn−−5) !( 5) ! =n ( n −1) ( n − 2) ( n − 3) ( n − 4)
89 (4 3n !− ) 7. Simplify the following factorial expression: . (4 2n !+ )
(43nn−− !43) ! ( ) = (42nnnnnnn+++−−− !42414414243) ( ) !( ) ( ) ( ) ( ) ( )
1 = (424144142nnnnn++−−) ( ) ( ) ( ) ( )
Determine whether each of the following series converges or diverges using the Ratio Test.
8. nn( !) n=1
Note that nn( !) has nonzero terms. n=1
Let an = n( n!) .
a (nn++11)( ! ) limlimn+1 = nn→→ ann n ( )( !) n +1 (n +1!) =lim n→ nn! n +1 (nn+1!) =lim n→ nn! nn++11 =lim n→ n 1 =
a Since lim n+1 =, the series diverges by the Ratio Test. n→ an
90 n2 9. n=1 (3!n)
n2 Note that has nonzero terms. n=1 (3!n)
n2 Let a = . n (3!n)
2 a (n +1) n2 limlimn+1 = nn→→ annn (33+ !3!) ( ) (nn+13!)2 ( ) =lim n→ (33nn+ ! ) 2 (3!1nn) ( + )2 =lim n→ (33nn+ ! ) 2
2 (3!n) n +1 =lim n→ (333231nnnnn+++)( 3! )( )( )
2 11n + =lim n→ (333231nnnn+++)( )( ) = 0
a Since lim0 n+1 = and 01 , the series converges by the Ratio Test. n→ an
91 23nn2 + 10. n n=3 4
23nn2 + Note that has nonzero terms. n n=3 4
23nn2 + Let a = . n 4n
2 a 2133(nn+++) 23nn2 + limlimn+1 = nn→→ nn+1 an 44
2 2133(nn+++) 4n =lim n→ 423n+12nn+
2 2133(nn+++) 4n =lim n→ 234nn21+ n+
2 2133(nn+++) 4n =lim n→ 234nn2 + 4 n
2 2133(nn+++) 1 =lim n→ 234nn2 + 1 = 4
a 1 1 Since lim n+1 = and 1, the series converges by the Ratio Test. n→ an 4 4
92 n 7 11. (n − 2) n=3 5
n 7 Note that (n − 2) has nonzero terms. n=3 5
n 7 Let ann =−( 2). 5
n+1 a ((n +−127) )( ) 5n limlimn+1 = nn→→ n+1 n an 5 (n − 27)( ) (n −1) 75nn+1 =lim n→ (n − 275) nn+1 (n −1) 7 75nn =lim n→ (n −275) 5 nn (n −1) 71 =lim n→ (n − 21) 5 7 = 5
a 7 7 Since lim n+1 = and 1, the series diverges by the Ratio Test. n→ an 5 5
We could have also determined that diverges using the Direct
n 7 Comparison Test and comparing it to the divergent geometric series . The Direct n=3 5 Comparison Test may be preferable since we can quickly identify that for n 3 nn 77 −(n 2) and therefore conclude that diverges without having 55 a to find lim n+1 using the Ratio Test. n→ an
93 For each of the following, match the series with the method you would use to determine convergence. Some series can be determined using more than one method. However, you may list each method only once.
1 12. C A. The nth-Term Test for Divergence 8 7 n=1 n (n + 2!) 13. F B. Geometric series n=1 11
8n 14. E C. p-series n n=1 13− 9
4 15. B D. Direct Comparison Test n n=1 3
92n − 16. A E. Limit Comparison Test n=1 71n +
6 17. D F. Ratio Test 2 n=1 n +1
94 Appendix: Guided Notes and Activity Sheets for Students
Lesson 1: Introduction to Sequences and Series Guided Notes ...... 96 Activity Sheet ...... 99
Lesson 2: The nth-Term Test for Divergence Guided Notes ...... 102 Activity Sheet ...... 108
Lesson 3: Geometric Series Guided Notes ...... 112 Activity Sheet ...... 118
Lesson 4: p-Series Guided Notes ...... 124 Activity Sheet ...... 134
Lesson 5: Direct Comparison Test Guided Notes ...... 138 Activity Sheet ...... 147
Lesson 6: Limit Comparison Test Guided Notes ...... 151 Activity Sheet ...... 161
Lesson 7: Ratio Test Guided Notes ...... 165 Activity Sheet ...... 175
95 Guided Notes
Introduction to Sequences and Series Name: ______
Definitions
Sequence:
We use subscript notation to represent the terms of the sequence: ______
The nth term is denoted by ____.
The entire sequence is denoted by ____.
Series: ______
We can calculate partial sums as well as some infinite sums.
______are denoted by Sm , where the first m terms of the sequence are added.
m Notation using sigma: Saaaaamnm==++++ 123 n=1
If the limit L of a sequence exists as n goes to infinity, then the sequence ______to L.
If the limit of a sequence does not exist as n goes to infinity, then the sequence ______.
3n Example 1: Write the first 5 terms of the sequence whose nth term is a = . n n!
96 Example 2: Find an expression for the nth term of the sequence 3 , 7 , 1 1 , 1 5 , . . . .
111 Example 3: Find an expression for the nth term of the sequence 1, ,−− , , ... . 2624
Example 4: Determine if the sequence converges or diverges by finding the limit, if 1 possible: a =−5. n n3
Example 5: Determine if the sequence converges or diverges by finding the limit, if 2 possible: an = cos. n
Example 6: Determine if the sequence converges or diverges by finding the limit, if n possible: an =1 +( − 1) .
97 Example 7: Determine if the sequence converges or diverges by finding the limit, if possible: ann = sin ( ) .
3n Example 8: Find the partial sum S given a = . 4 n n + 2
98 Activity Sheet
Activity: Introduction to Sequences and Series Name: ______
1. Classify each of the following as a sequence or a series.
1 1 1 1 4 8 16 32 a) , , , , ... b) + + + + 2345 5 125 625 3125
12624 3333 c) ++++ d) −−, , , , ... 7777 8163264
2. Create your own pattern for a sequence. Write the expression for the nth term, using correct notation. Then list the first five terms of your sequence.
3. For each of the following, write an expression for the nth term of the sequence.
5 25125 6253125 a) −−−, , , , , ... 33333
27999 b) 27, , , , , ... 22840
99 c) 1, 6, 11, 16, 21, ...
2624120720 d) −−−−−, , , , , ... 749343240116807
Find the first five terms of each of the following sequences. Then determine if the sequence converges or diverges. If the sequence converges, find its limit.
5n 4. a = n 3+ n
n 5. ann =−( 1!)
100 5n 6. a = n n3
21 7. Find the partial sum S6 of the series −+−+−+−+42( ) 36
101 Guided Notes
The nth-Term Test for Divergence Name: ______
Recall from previous lesson
Series: ______
We can calculate partial sums as well as some infinite sums.
m Notation using sigma: Saaaaamnm==++++ 123 n=1
Example 1: Consider the series 0 0++++ 0 0 .
th If we let an represent the n term of this series, what is l im an ? n→
If we added infinitely many terms of this series, what do you think the sum would be?
Based on our answers above, do you think this series converges or diverges?
111111111 Example 2: Consider the series 1 ++++++++++. 223334444
If we let represent the nth term of this series, what is ?
If we added infinitely many terms of this series, what do you think the sum would be?
102 Based on our answers above, do you think this series converges or diverges?
Example 3: Consider the series 5791113+++++ .
th If we let an represent the n term of this series, what is l im an ? n→
If we added infinitely many terms of this series, what do you think the sum would be?
Based on our answers above, do you think this series converges or diverges?
11111 Example 4: Consider the series +++++ . 33333
If we let represent the nth term of this series, what is ?
If we added infinitely many terms of this series, what do you think the sum would be?
Based on our answers above, do you think this series converges or diverges?
103 Based on our examples, what conclusion can we make regarding l im an and the n→ convergence or divergence of a series?
This leads us to our first test to determine whether a series diverges.
Think about the following:
If a series converges, the limit of its nth term must be 0.
That is, if a converges, then l im 0an = . n n→ n=1
It’s generally easier to determine the limit of the nth term than it is to determine whether a series converges. Therefore, we will use the ______of the statement above.
The nth-Term Test for Divergence
If lim0an , then an diverges. n→
Note: If , we cannot make any conclusion about the series. Look back at Example 1 and Example 2 to verify this.
Additionally, notice that when we express a sum as _____ without ______, we always mean ______, starting at a finite value of n. Since the series has ______
_____, the convergence or divergence of the series does not depend on the ______
______. However, the ______will depend on the initial value of n.
104 For example, let’s look at aaaann=++12 . Since a1 and a2 are ______, nn==13 an ______if and only if an ______. n=1 n=3
Example 5: Use the nth-Term Test for Divergence to determine whether the series diverges.
n 5 1( .2 3 ) n=0
Example 6: Use the nth-Term Test for Divergence to determine whether the series diverges.
7 n=3 nn( + 8)
105 Example 7: Use the nth-Term Test for Divergence to determine whether the series diverges.
n n=1 23n +
Example 8: Use the nth-Term Test for Divergence to determine whether the series diverges.
3n n=1 n!
106 Example 9: Use the nth-Term Test for Divergence to determine whether the series diverges.
1 9 − 3 n=0 n
107 Activity Sheet
Activity: The nth-Term Test for Divergence Name:______
Determine if each of the following statements are True or False.
1. If l im 0an = , then an must converge. n→
2. If diverges, then .
3. If converges, then .
4. If , then must diverge.
5. If lim0an , then must converge. n→
6. If , then must diverge.
7. If diverges, then .
8. If converges, then lim0an . n→
Based on your results above, only statements _____ and _____ are true.
Therefore, if we can show that______, then we can conclude that a series ______using the nth-Term Test for Divergence.
If we can show that______, then the nth-Term Test for Divergence is ______.
108 8 9. Use a table to find l im an given a = . Round your answers to the nearest n→ n n + 9 hundred thousandth.
n 1 10 100 1,000 10,000 100,000
an 8 Based on your table, l im an = ____. What conclusion can we make about ? n→ n + 9
(1)!n + 10. Use a table to find given a = . n n!
1 10 100 1,000 10,000 100,000
(n +1!) Based on your table, ____. What conclusion can we make about ? n!
th For each of the following, use algebraic techniques to find lim an . Then apply the n - n→ Term Test if possible.
7n 11. n=1 31n −
109 −+32n 12. 2 n=4 41n −
n 1 13. 3 n=0 4
−+52nn3 14. 2 n=2 891nn+−
110 15. 7 n=1
6 16. −+5 n=3 n
n 3n 17. (−1) n=4 56n −
111 Guided Notes
Geometric Series Name: ______
3333 Consider the series 3 +++++. 24816
What do you notice about the terms of this series?
Series consisting of terms with a constant ratio are called ______. The terms of this type of series form a ______.
In general, the series given by araararararnn=++++++ 23 with a 0 and n=0 r 0 is a ______with ______r.
Example 1: Identify the ratio of the geometric series and write the series in sigma notation.
5+ 10 + 20 + 40 + 80 +
Example 2: Identify the ratio of the geometric series and write the series in sigma notation.
8 2 1 1 1 + + + + + 5 5 10 40 160
112 Example 3: Identify the ratio of the geometric series and write the series in sigma notation.
1392781+−++−++( ) ( )
Identifying the ratio of a geometric series is a crucial skill because the ratio determines whether the series will converge or diverge. Let’s investigate this below.
You may remember from Algebra 2 that there is a formula for the sum of the terms of a m finite geometric sequence: ar n = ______provided that ______. n=0 m If r =1, then ar n = ______. n=0
We know that the value of an infinite series is the limit of its partial sums.
Using the formula above for r 1, we can represent the mth partial sum as
______. Now, let’s take the limit of this partial sum.
If we want this sequence of partial sums to converge, we want to find what values of r will make ______exist.
If ______or ______, lim r m does not exist. m→
When ______, ______and the sequence of partial sums ______.
113 So assuming r 1, we can find l im Sm . m→
m n The other possibility from above was r =1 and S arm == ______. n=0
In this case, l im Sm = ______and therefore this geometric series would diverge. m→
Thus a geometric series with ratio r ______if and will have the sum
ar n = ______. n=0
• Notice that this characterization for the limit of the sum requires the value of the index to start at ______. We will address how to calculate sums when the index starts at other values in the examples to follow.
A geometric series with ratio r ______if r 1.
Example 4: Determine whether the geometric series converges or diverges. If the series converges, find the sum.
n 2 3 n=0 7
114 Example 5: Determine whether the geometric series converges or diverges. If the series converges, find the sum.
6
n n=1 8
115 Example 6: Determine whether the geometric series converges or diverges. If the series converges, find the sum.
n 6 − n=2 11
116 Example 7: Determine whether the geometric series converges or diverges. If the series converges, find the sum.
1 n − (8) n=0 5
Example 8: Determine whether the geometric series converges or diverges. If the series converges, find the sum.
en n=1 4
117 Activity Sheet
Activity: Geometric Series Name: ______
1. Determine if each of the following is a geometric series. If it is a geometric series, write the series in summation notation.
404040 a) 12040+++++ b) −+−++++832712( ) 3927
c) −++−++−+2481632( ) ( ) d) + 2345 + + + +
251256253125 e) −+++++1763215511 f) 10 +++++ 2832128
125 125 125 g) 1491625+++++ h) 1000+( − 250) + + − + + 2 8 32
118 Determine if each of the following is a geometric series. If it is a geometric series, identify the ratio r.
n 3 2 n 2. 5− 3. n=0 3 n=0 5
n 19n 4. −34( ) 5. n=1 n=0 24
7 3n4 6. n 7. n n=3 13 n=2 6
n! 51n − 8. n 9. 2 n=2 43+ n=0 842nn−+
1 n n3 10. ( ) 11. n=1 3 n=0 79n −
119 12. A geometric series diverges if ______.
13. A geometric series converges if ______.
14. For a convergent geometric series, ar n = ______. n=0
15. Describe two ways to calculate the sum of a convergent geometric series ar n . n=3
Determine whether each of the following geometric series converges or diverges. If the series converges, find the sum.
n 1 16. 4 n=0 7
120 n 7 17. 5 n=0 3
5 18. n n=1 6
121 3 n 19. − ( 11) n=0 4
n 5 20. −1 n=2 6
122 n 2 21. − n=0 9
1 n 22. ( 7 ) n=1 8
23. Write a geometric series in summation notation that converges to a sum of 6.
123 Guided Notes p-Series Name: ______
1111 Example 1: Consider the series 1 +++++. 491625
What pattern do you notice in the terms of this series?
What term does not seem to fit this pattern at first glance?
Can you verify that this term matches the pattern we notice?
3133 Example 2: Consider the series 3 +++++. 8964125
What pattern do you notice in the terms of this series?
What terms do not seem to fit this pattern at first glance?
Can we verify that these terms match the pattern we notice?
124 11111 A series of the form ppppp=++++ is a ______where p is ______n=1 n 1234 ______.
11111 Additionally, when ______, the series =++++ is called the ______n=1 n 1234 ______.
The series in Example 1 and Example 2 are both p-series.
1111 We can rewrite Example 1 as 1 +++++= ______where ______. 491625
It is important to notice that Example 2 has a numerator other than 1. However, we know that infinite series are limits of finite sums. Based on properties of limits that we have previously studied, we know we can factor constants out of limits. Therefore, we can also factor constants out of infinite series.
kakann= where k is a constant. nn==11
3133 Therefore, Example 2 can be rewritten as 3 +++++= ______. 8964125
Example 3: Identify the value of p for the p-series and write the series in sigma notation.
1111 1+++++ 32243 10243125
125 Example 4: Identify the value of p for the p-series and write the series in sigma notation.
1111 1+++++ 2345
Example 5: Identify the value of p for the p-series and write the series in sigma notation.
7777 7+++++ 44234544
Example 6: Identify the value of p for the p-series and write the series in sigma notation.
11111111 11 +++++ 223344553333
126 Like finding the common ratio in a geometric series, identifying the value of p in a p- series is a crucial skill because p determines whether the series will converge or diverge.
A p-series with p > 1 will ______.
A p-series with 0 < p < 1 will ______.
Remember that p must be ______.
Let’s investigate why the convergent and divergent statements above are true.
Remember that an infinite series is a sequence of partial sums. When we determine 1 whether converges, we look at the sequence of partial sums. p n=1 n
Let’s begin by looking at the partial sums of .
127 Think back to earlier in this course. When have we added up values of a function?
The partial sums we have listed above look different from Riemann sums since each term is not being multiplied by x . However, if we let =x 1, we can rewrite the partial sums as follows:
Now, we can see that each of these partial sums can be interpreted as a ______.
m 1 Let’s look at Riemann sum approximations for dx . p 1 x
A left Riemann sum approximation with =x 1 would be:
m 1 dx ______. p 1 x
128 This would correspond to the partial sum ______.
1 We know that ______and thus fx()= is a ______function on the interval _____. x p
Left Riemann sum approximation [4].
Therefore, the left Riemann sum approximation is an ______of the integral.
m 1 That is, S dx . m−1 p 1 x
For 01p ,
m 1 limSm−1 lim dx mm→ → p 1 x m x(−+p 1) = lim m→ −+p 1 1 m(−pp +11) 1( − + ) =−lim . m→ −pp +11 − +
When , −+p 1 is positive and therefore m(−+p 1) grows without bound as m(−pp +11) 1( − + ) m →. Thus lim − is infinite. This implies that lim Sm−1 is ______ m→ m→ −pp +11 − + 1 and the sequence of partial sums ______. Thus, ______. p n=1 n
129 For p =1,
m 1 l im lS im dxm−1 mm→→ 1 1 x m = l im l n x m→ 1 =−l im l n l n 1m m→ ( ) =−l im l n 0 m m→ ( ) =.
Thus l im Sm−1 is ______and therefore the sequence of partial sums ______. Thus, m→ 1 ______. p n=1 n
A right Riemann sum approximation with =x 1 would be:
m 1 dx ______. p 1 x
This would correspond to the partial sum ______.
1 We know that ______and thus fx()= is a ______function on the interval _____. x p
Right Riemann sum approximation [4].
130 Therefore, the right Riemann sum approximation is an ______of the integral.
11m 11m That is, S−1 dx , so S+1 dx . m pp( ) m pp( ) 1 1 x 1 1 x
For p 1,
11m limlim1Sdxm +( ) mm→→ pp 1 1 x 11m =+(1lim) dx ppm→ 1 1 x m 1 x(−+p 1) =+(1lim) 11p m→ −+p 1 11m(−pp +−11) + ( ) =+− (1lim.) p m→ 111 −+−+pp
11m(−pp +−11) + ( ) Thus, lim1lim.S +− m p ( ) mm→→ 111 −+−+pp
When p 1, −+p 1 is negative and therefore m(−+p 1) converges to 0 as m →. Thus 11m(−pp +11) ( − + ) (1) +− lim ______. This implies that l im Sm ______. Since the limit p m→ 1m→ −pp + 1 − + 1 1 of partial sums ______, ______. p n=1 n
Therefore, converges when ______and diverges when ______.
We will usually assign p-series an initial index value of n =1, however the initial value of the index can be any finite positive value. This will not affect the convergence of a p- series.
131 Using this information, we can determine whether the series in Examples 3 through 6 converge.
Example 3:
Example 4:
Example 5:
Example 6:
Example 7: Determine whether the p-series converges or diverges.
11111 6/76/76/76/76/7=+++++ 1 n=1 n 2345
Example 8: Determine whether the p-series converges or diverges.
9 9 9 9 9 = 9 + + + + + n=1 n 2345
132 Example 9: Determine whether the p-series converges or diverges.
3 3 3 3 3 = 3 + + + + + 666 6 6 n=1 n 2345
Example 10: Determine whether the p-series converges or diverges.
8.38.38.38.38.3 =+++++ 8.3 n=1 n 2345
133 Activity Sheet
Activity: p-Series Name: ______
1. Determine if each of the following is a p-series. If it is, write the p-series in summation notation.
1111 7777 a) 1+++++ b) 7 +++++ 2345−−−−1111 2345
1234 17171717 5555 c) 17 +++++ d) 77777+++++ 223344559999 2222
−5 − 5 − 5 − 5 − 5 1414 e) 1+ 2 + 3 + 4 + 5 + f) 4 +++++ 22716125
134 Identify p in each of the following p-series. Then, determine if the series converges or diverges.
1 1 2. 3/7 3. e n=1 n n=1 n
1 4. Write a p-series in summation notation with p = and whose first term is 9. 45
5. Write a p-series in summation notation with p = 6 and whose first term is 1.
Determine whether each of the following series converges or diverges. Use any method we have studied.
83+ n n 6. 7. 53( ) n=1 9n n=2
135 13 −2n 8. 9. 2 5 n=1 nn n=0 3
5.2n −3 10. 3 11. 8 n=2 n n=1 n
n 1 4 12. 13. n=0 6 n=1 n
136 4n 16.2 14. 15. n n=3 n=0 35
16. Complete the following chart.
General form of Series diverges if … Series converges if… series
The nth-Term Test for
Divergence
Geometric Series
p-Series
17. The nth-Term Test for Divergence can only determine that a series ______.
18. We can calculate the sum of a convergent ______with an initial index value of ______using the formula ______.
1 19. In a ______in the form p , ______must be a ______. nm= n
137 Guided Notes
Direct Comparison Test Name: ______
1 Example 1: We are given the series and asked to determine whether the series 2 n=1 n +1 converges or diverges. What methods might we try in order to answer this question?
1.
2.
3.
None of the methods we have studied so far is applicable to this particular series.
What if instead of matching a known series type exactly, we looked for a type that was similar to the series we are working with?
138 1 The series is similar to ______. 2 n=1 n +1
What ______would we say is most similar to ?
What do we know about the series we wrote above?
Let’s list a few terms of the original series we were given and the new series we have compared it to.
Let’s compare the two series term by term.
We can see that each term of the original series is ______the corresponding term of the series we selected.
Additionally, we can compare the expression for the nth term for each series.
139 What do we notice about these expressions?
1 Since we know that the series ______, what do you think this means about 2 n=1 n our original series?
This leads us to the next convergence test we will study: the Direct Comparison Test.
Direct Comparison Test
Given two series an and bn , let 0 abnn for all n. n=1 n=1
1. If bn converges, then an converges. n=1 n=1
2. If diverges, then diverges.
To use this test, we must find a series similar to the series we are given. We should choose a series that is either ______or ______since we can easily determine whether these series converge or diverge.
140 Additionally, notice that this test requires both an and bn to consist of positive n=1 n=1 terms. This is an important condition to verify before applying the Direct Comparison Test.
Let’s investigate why this condition is important. We will assume we only need abnn .
−1 1 a = Let n and bn = 2 . nn==11n nn==11n
It is true that for all terms since every an is negative and every bn is positive. We −1 1 know that ______since it is the negative of the ______and is a 2 n=1 n n=1 n ______that ______. So even with , when the terms of the series are not all
positive, we cannot use the convergence of bn to determine the convergence of an . n=1 n=1
Also even with , we cannot use divergence of to determine the divergence of .
Therefore, we must be sure that we are working with series consisting of positive terms when we apply this test.
141 Example 2: Use the Direct Comparison Test to determine whether the series converges or diverges.
5
3 n=1 n
142 Example 3: Use the Direct Comparison Test to determine whether the series converges or diverges.
2n n n=1 35+
143 Example 4: Use the Direct Comparison Test to determine whether the series converges or diverges.
8 n=1 n + 3
144 Example 5: Use the Direct Comparison Test to determine whether the series converges or diverges.
1
5 n=1 n +1
145 Example 6: Use the Direct Comparison Test to determine whether the series converges or diverges.
1
4 n=1 72n −
146 Activity Sheet
Activity: Direct Comparison Test Name: ______
Determine if each of the following is similar to a geometric series or a p-series. Then provide the series you would use with the Direct Comparison Test.
52n + 1 1. 2. n 3 n=1 4 n=1 31n +
1 6n 3. 4. 7 5 n n=1 109 n − n=1 117+
Use the words “smaller” and “larger” to complete the following statements describing the results of the Direct Comparison Test.
5. If the ______series converges, the ______series must also converge.
6. If the ______series diverges, the ______series must also diverge.
147 Determine whether each of the following series converges or diverges using the Direct Comparison Test.
2 7. 6 n=1 n +12
148 8n 8. n n=1 74−
149 0 ab 9. Given two series an and bn with nn, describe two scenarios that are n=1 n=1 inconclusive when using the Direct Comparison Test.
State the method you would use to determine the convergence of each of the following series. You do not need to determine whether the series converges or diverges.
6 9 10. 11. 5 3 n n=1 n n=1 5
2n 118n − 12. n 13. n=1 37+ n=1 57n +
150 Guided Notes
Limit Comparison Test Name: ______
7 Example 1: We are given the series and asked to determine whether the series n n=1 65− converges or diverges. What methods might we try in order to answer this question?
Note: Our instincts should lead us to the Direct Comparison Test, but let’s investigate the other methods we have learned as well.
1.
2.
3.
4.
151
None of the methods we have studied so far is applicable to this particular series.
However, there is another test we can use to determine the convergence or divergence of a series: The Limit Comparison Test.
As in the Direct Comparison Test, the Limit Comparison Test will require us to use a series that is similar to the series we are given.
152 Limit Comparison Test
an Suppose that an 0 , bn 0 , and l im = L where L is finite and positive. Then the n→ bn two series an and bn either both converge or both diverge.
Let’s take a look at the possibilities for L to see why L must be finite and positive in order to apply the Limit Comparison Test.
Since we must begin with and , we know that the limit of the ratio of these terms must be nonnegative if the limit exists. Therefore, we know L 0 .
Case 1: If L is a ______value, we can conclude that the limit of the ratio of terms of ______is this same positive finite value. Therefore when we look far enough out in the series, the terms of are about ______.
So the partial sums for converge if and only if ______.
Likewise, the partial sums for diverge if and only if ______.
Case 2: If L = 0 or the limit does not exist, the Limit Comparison Test is inconclusive. The following examples demonstrates this.
1 1 Let’s use the series 2 and . Both series consist of positive terms. n=1 n +1 n=1 n
153 a Now, find l im n . n→ bn
Let’s change the labels and find once again.
We know ______by the ______(see Lesson 5) and we know that ______is the ______which ______. So in these examples, when
______or ______, one series converges while the other diverges.
Note: There are other examples where both series converge or both series diverge even when L is not finite and positive.
To summarize, we have shown that we can only apply the Limit Comparison Test to two series consisting of positive terms if the limit of the ratio of their terms is a ______.
154 7 Going back to Example 1, we can compare to ______. When we were n n=1 65− trying to apply the Direct Comparison Test, we used ______as the comparison series but were unable to apply the test because ______has smaller terms than .
Since we do not need to decide what series has larger or smaller terms in order to use the Limit Comparison Test, we will use the simpler version of the geometric series in the comparison.
The series ______is a ______with ______and ______. Therefore,
______.
Note that both the original series ______and new series ______consist of positive terms.
Let’s apply the Limit Comparison Test.
a Now, find lim n . n→ bn
155 What if we interchanged the labels before calculating the limit?
a Find lim n . n→ bn
Notice that it did not matter which series we labeled as an or bn . Both limits produced positive, finite values.
When using the Limit Comparison Test, we can choose to label the given series as either or . In some cases, we may find that changing how we label the series may help us simplify the limit.
However, to be consistent in these notes, we will label the given series as .
156 Example 2: Use the Limit Comparison Test to determine whether the series converges or diverges.
1 n=1 n + 4
157 Example 3: Use the Limit Comparison Test to determine whether the series converges or diverges.
3nn 5 12 +− 73 n=1 4 8nn 9++
158 Example 4: Use the Limit Comparison Test to determine whether the series converges or diverges.
5n n n=0 27+
159 Example 5: Use the Limit Comparison Test to determine whether the series converges or diverges.
3n n n=1 41−
160 Activity Sheet
Activity: Limit Comparison Test Name: ______
a 1. The Limit Comparison Test requires us to calculate l im n = L . What must be true n→ bn about L in order to be able to apply this test?
Determine whether each of the following series converge or diverge using the Limit Comparison Test.
5 2. 4 3 n=1 n + 8
161 84n + 3. n n=0 1 3 7+
162 8nn 3 72 +− 4. 3 n=1 92nn−
5. When asked to determine if a series converges or diverges, what would lead you to try the Limit Comparison Test rather than other methods?
163 Let’s assume an 0 and bn 0 . You have done the work to calculate the following limits. If you arrived at each of the following conclusions, would you be able to apply the Limit Comparison Test? Why or why not?
a 6. l im n = n→ bn
a 1 7. l im n = n→ bn 3
a 8. lim1 n = n→ bn
a 9. lim0 n = n→ bn
a 10. lim7 n = n→ bn
164 Guided Notes
Ratio Test Name: ______
n Example 1: We are given the series and asked to determine whether the series n n=1 5 converges or diverges. What methods might we try in order to answer this question?
1.
2.
3.
4.
165 None of the methods we have studied so far is applicable to this particular series.
However, there is another test we can use to determine the convergence or divergence of a series: The Ratio Test.
The Ratio Test is useful since we do not need to compare the given series to another series. We only need to use the given series and the expression for its terms. We can also use the Ratio Test for series with negative terms.
Ratio Test
Let an be a series with nonzero terms.
a 4. The series converges if l im 1 n+1 . n→ an a a 5. The series diverges if lim1 n+1 or lim n+1 =. n→ n→ an an a a 6. The Ratio Test is inconclusive if lim1 n+1 = or lim n+1 does not exist. n→ n→ an an
The absolute values are important, especially for series with some positive terms and some negative terms, but where the positive and negative values do not alternate from one term to the next. In a case such as this, the limit of the ratio of successive terms with the absolute value would exist, but the limit without the absolute value would not exist. We will not investigate a case such as this in the notes but interested students can ask to see an example later.
166 a The Ratio Test requires us to find l im n+1 . Unlike the Limit Comparison Test, we n→ an cannot calculate the limit of the reciprocal and arrive at the same conclusion. For a a 1 example, if l im 5 n+1 = , then l im n = . This would lead to two different n→ n→ an an+1 5 conclusions using the Ratio Test.
n Let’s apply the Ratio Test to the series in Example 1: . n n=1 5
167 Example 2: Use the Ratio Test to determine whether the series converges or diverges.
n 2 n n=1 3
168 Example 3: Use the Ratio Test to determine whether the series converges or diverges.
8n 6 n=2 n
169 Example 4: Use the Ratio Test to determine whether the series converges or diverges.
6 n=1 n
170 Example 5: Use the Ratio Test to determine whether the series converges or diverges.
3
4 n=2 n +1
171 Example 6: Use the Ratio Test to determine whether the series converges or diverges.
9n n=2 n!
172 Example 7: Use the Ratio Test to determine whether the series converges or diverges.
(n −1!) 7 n=1 n
173 Example 8: Use the Ratio Test to determine whether the series converges or diverges.
n −9 n n=1 7
174 Activity Sheet
Activity: Ratio Test Name: ______
Complete the following statements regarding the Ratio Test.
1. The series an must have ______terms.
a 2. If l im 1 n+1 , then ______. n→ an
a a 3. If l im 1 n+1 or l im n+1 =, then ______. n→ n→ an an
a 4. If lim1 n+1 = , ______. n→ an
5. Describe an advantage of using the Ratio Test instead of the Direct or Limit Comparison Tests.
n! 6. Simplify the following factorial expression: . (n − 5!)
175 (4 3n !− ) 7. Simplify the following factorial expression: . (4 2n !+ )
Determine whether each of the following series converges or diverges using the Ratio Test.
8. nn( !) n=1
176 n2 9. n=1 (3!n)
177 23nn2 + 10. n n=3 4
178 n 7 11. (n − 2) n=3 5
179 For each of the following, match the series with the method you would use to determine convergence. Some series can be determined using more than one method. However, you may list each method only once.
1 12. A. The nth-Term Test for Divergence 8 7 n=1 n (n + 2!) 13. B. Geometric series n=1 11
8n 14. n C. p-series n=1 1 3 9−
4 15. D. Direct Comparison Test n n=1 3
92n − 16. E. Limit Comparison Test n=1 71n +
6 17. F. Ratio Test 2 n=1 n +1
180 References
[1] AP Central. (2019). https://apcentral.collegeboard.org/pdf/ap-calculus-ab-and-bc- course-and-exam-description.pdf. Accessed 2 June 2019.
[2] Larson, R., Hostetler, R.P., & Edwards, B.H. (2006). Calculus of a Single Variable. 8th Edition. Houghton Mifflin Company.
[3] Quora. (2018). https://www.quora.com/Whats-an-intuitive-proof-of-the-limit- comparison-test-I-havent-been-able-to-understand-the-rigorous-proofs-and-its-really- annoying-me-that-I-cant-seem-to-grasp-exactly-WHY-the-test-works. Accessed 17 July 2019.
[4] Texas Instruments (2019). Calculus: Riemann Sums. https://education.ti.com/en/timathnspired/us/detail?id=3CD2788D70CE46C6837BED58 EA69C255&sa=B03658FB92E74FA8A90BA1F7381D9FCD&t=2596C50B247541C1B 54AC803ECC87BD1. Accessed 18 July 2019.
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